How do I find the Vertex of the Parabola from General Form? – Recall, that the General Form of a parabola is y = ax 2 + bx + c. In order to find the vertex from this form, you must first find the x-coordinate of the vertex which is x = – b/2a. After you find the x-coordinate of the vertex, you will take this number and substitute for x in the parabola equation.

#### What is the formula for the vertex?

What Is Vertex Form? – While the standard quadratic form is \$ax^2+bx+c=y\$, the vertex form of a quadratic equation is \$\bi y=\bi a(\bi x-\bi h)^2+ \bi k\$. In both forms, \$y\$ is the \$y\$-coordinate, \$x\$ is the \$x\$-coordinate, and \$a\$ is the constant that tells you whether the parabola is facing up (\$+a\$) or down (\$-a\$).

I think about it as if the parabola was a bowl of applesauce; if there’s a \$+a\$, I can add applesauce to the bowl; if there’s a \$-a\$, I can shake the applesauce out of the bowl.) The difference between a parabola’s standard form and vertex form is that the vertex form of the equation also gives you the parabola’s vertex: \$(h,k)\$.

For example, take a look at this fine parabola, \$y=3(x+4/3)^2-2\$: Based on the graph, the parabola’s vertex looks to be something like (-1.5,-2), but it’s hard to tell exactly where the vertex is from just the graph alone. Fortunately, based on the equation \$y=3(x+4/3)^2-2\$, we know the vertex of this parabola is \$(-4/3,-2)\$.

Why is the vertex \$(-4/3,-2)\$ and not \$(4/3,-2)\$ (other than the graph, which makes it clear both the \$x\$- and \$y\$-coordinates of the vertex are negative)? Remember: in the vertex form equation, \$h\$ is subtracted and \$k\$ is added, If you have a negative \$h\$ or a negative \$k\$, you’ll need to make sure that you subtract the negative \$h\$ and add the negative \$k\$.

In this case, this means: \$y=3(x+4/3)^2-2=3(x-(-4/3))^2+(-2)\$ and so the vertex is \$(-4/3,-2)\$. You should always double-check your positive and negative signs when writing out a parabola in vertex form, particularly if the vertex does not have positive \$x\$ and \$y\$ values (or for you quadrant-heads out there, if it’s not in quadrant I ).

1. This is similar to the check you’d do if you were solving the quadratic formula (\$x= }/ \$) and needed to make sure you kept your positive and negatives straight for your \$a\$s, \$b\$s, and \$c\$s.
2. Below is a table with further examples of a few other parabola vertex form equations, along with their vertices.

Note in particular the difference in the \$(x-h)^2\$ part of the parabola vertex form equation when the \$x\$ coordinate of the vertex is negative.

 Parabola Vertex Form Vertex Coordinates \$y=5(x-4)^2+17\$ \$(4,17)\$ \$y=2/3(x-8)^2-1/3\$ \$(8,-1/3)\$ \$y=144(x+1/2)^2-2\$ \$(-1/2,-2)\$ \$y=1.8(x+2.4)^2+2.4\$ \$(-2.4,2.4)\$

#### How do you find H and K in vertex form?

Find h, the x-coordinate of the vertex, by substituting a and b into h=–b2a. Find k, the y-coordinate of the vertex, by evaluating k=f(h)=f(−b2a).

#### What is the formula for a parabola?

The standard form of a parabola is y=ax^2+bx+c where a, b, and c are real numbers and a is not equal to zero.

## Is vertex the same as turning point?

How to find the turning point of a parabola?

• Improve Article
• Save Article
• Like Article

When a quadratic equation is represented graphically with a U – shape, it is called a parabola. A parabola can also be defined as a plain curve where any point on that curve is equidistant from a fixed point, the focus. The turning point of any curve or parabola is the point at which its direction changes from upward to downward or vice-versa.

## Is H or K the vertex?

(finding the equation of a parabola) h: is the horizontal coordinate of the vertex. k: is the vertical coordinate of the vertex.

## What is H in parabola?

 The vertex form of a quadratic function is given by f ( x ) = a ( x – h ) 2 + k, where ( h, k ) is the vertex of the parabola.

FYI: Different textbooks have different interpretations of the reference ” standard form ” of a quadratic function. Some say f ( x ) = ax 2 + bx + c is “standard form”, while others say that f ( x ) = a ( x – h ) 2 + k is “standard form”. To avoid confusion, this site will not refer to either as “standard form”, but will reference f ( x ) = a ( x – h ) 2 + k as “vertex form” and will reference f(x ) = ax 2 + bx + c by its full statement.

When written in ” vertex form “: • (h, k) is the vertex of the parabola, and x = h is the axis of symmetry. • the h represents a horizontal shift (how far left, or right, the graph has shifted from x = 0). • the k represents a vertical shift (how far up, or down, the graph has shifted from y = 0). • notice that the h value is subtracted in this form, and that the k value is added.

If the equation is y = 2( x – 1) 2 + 5, the value of h is 1, and k is 5. If the equation is y = 3( x + 4) 2 – 6, the value of h is -4, and k is -6. To Convert from f ( x ) = ax 2 + bx + c Form to Vertex Form: Method 1: Completing the Square To convert a quadratic from y = ax 2 + bx + c form to vertex form, y = a ( x – h ) 2 + k, you use the process of completing the square. Let’s see an example. Convert y = 2 x 2 – 4 x + 5 into vertex form, and state the vertex.

 Equation in y = ax 2 + bx + c form. y = 2 x 2 – 4 x + 5 Since we will be ” completing the square ” we will isolate the x 2 and x terms, so move the + 5 to the other side of the equal sign. y – 5 = 2 x 2 – 4 x We need a leading coefficient of 1 for completing the square, so factor out the current leading coefficient of 2. y – 5 = 2( x 2 – 2 x ) Get ready to create a perfect square trinomial. BUT be careful!! In previous completing the square problems with a leading coefficient not 1, our equations were set equal to 0. Now, we have to deal with an additional variable, “y”, so we cannot “get rid of ” the factored 2. When we add a box to both sides, the box will be multiplied by 2 on both sides of the equal sign. Find the perfect square trinomial. Take half of the coefficient of the x -term inside the parentheses, square it, and place it in the box, Simplify and convert the right side to a squared expression. y – 3 = 2( x – 1) 2 Isolate the y -term, so move the -3 to the other side of the equal sign. y = 2( x – 1) 2 + 3 In some cases, you may need to transform the equation into the “exact” vertex form of y = a ( x – h ) 2 + k, showing a “subtraction” sign in the parentheses before the h term, and the “addition” of the k term. (This was not needed in this problem.) y = 2( x – 1) 2 + 3 Vertex form of the equation. Vertex = ( h, k ) = (1, 3) (The vertex of this graph will be moved one unit to the right and three units up from (0,0), the vertex of its parent y = x 2,)

Here’s a sneaky, quick tidbit: Method 2: Using the “sneaky tidbit”, seen above, to convert to vertex form:

 y = ax 2 + bx + c form of the equation. y = 2 x 2 – 4 x + 5 Find the vertex, ( h, k ), and, a = 2 and b = -4 Vertex: (1,3) Write the vertex form. y = a ( x – h ) 2 + k y = 2( x – 1) 2 + 3

To Convert from Vertex Form to y = ax 2 + bx + c Form:

 Simply multiply out and combine like terms: y = 2( x – 1) 2 + 3 y = 2( x 2 – 2 x + 1) + 3 y = 2 x 2 – 4 x + 2 + 3 y = 2 x 2 – 4 x + 5

Graphing a Quadratic Function in Vertex Form:

 1. Start with the function in vertex form : y = a ( x – h ) 2 + k y = 3( x – 2) 2 – 4 2. Pull out the values for h and k, If necessary, rewrite the function so you can clearly see the h and k values. ( h, k ) is the vertex of the parabola. Plot the vertex. y = 3( x – 2) 2 + (-4) h = 2; k = -4 Vertex: (2, -4) 3. The line x = h is the axis of symmetry, Draw the axis of symmetry. x = 2 is the axis of symmetry 4. Find two or three points on one side of the axis of symmetry, by substituting your chosen x -values into the equation. For this problem, we chose (to the left of the axis of symmetry): x = 1; y = 3( 1 – 2) 2 – 4 = -1 x = 0; y = 3( 0 – 2) 2 – 4 = 8 Plot (1, -1) and (0,8) 5. Plot the mirror images of these points across the axis of symmetry, or plot new points on the right side. Draw the parabola. Remember, when drawing the parabola to avoid “connecting the dots” with straight line segments. A parabola is curved, not straight, as its slope is not constant.

NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered “fair use” for educators. Please read the ” Terms of Use “.

/td>

#### Is the vertex the end point?

What Is an Endpoint in Geometry? – An endpoint is defined as a point at which a line segment or a ray ends. In general, an endpoint is the furthermost or the ending point. Examples: In a line segment, endpoints are the points at which the line segment ends. A line segment has two endpoints. A ray has one endpoint. In an angle, the common point between the two rays (vertex) is an endpoint. A polygon is formed using three or more line segments. Each side intersects two other sides, at an endpoint. In polygons, the points that join the sides are endpoints.

#### What is P in a parabola?

A parabola is a conic section, It is a slice of a right cone parallel to one side (a generating line) of the cone. Like the circle, the parabola is a quadratic relation, but unlike the circle, either x will be squared or y will be squared, but not both. You worked with parabolas in Algebra 1 when you graphed quadratic equations. We will now be investigating the conic form of the parabola equation to learn more about the parabola’s graph.

 A parabola is defined as the set ( locus ) of points that are equidistant from both the directrix (a fixed straight line) and the focus (a fixed point).

table>

This definition may be hard to visualize. Let’s take a look. For ANY point on a parabola, the distance from that point to the focus is the same as the distance from that point to the directrix. (Looks like hairy spider legs!) Notice that the “distance” being measured to the directrix is always the shortest distance (the perpendicular distance). The specific distance from the vertex (the turning point of the parabola) to the focus is traditionally labeled ” p “. Thus, the distance from the vertex to the directrix is also ” p “. The focus is a point which lies “inside” the parabola on the axis of symmetry. The directrix is a line that is ⊥ to the axis of symmetry and lies “outside” the parabola (not intersecting with the parabola).

Conic Equations of Parabolas: You recognize the equation of a parabola as being y = x 2 or y = ax 2 + bx + c from your study of quadratics. And, of course, these remain popular equation forms of a parabola. But, if we examine a parabola in relation to its focal point (focus) and directrix, we can determine more information about the parabola.

 Parabola with Vertex at Origin (0,0) (axis of symmetry parallel to the y -axis) Conic Forms of Parabola Equations: with the vertex at (0,0), focus at (0, p ) and directrix y = -p In the example at the right, the coefficient of x ² is 1, so, making p = ¼. The vertex is (0,0), the focus is (0,¼), and the directrix is y = -¼. The distance from the vertex (in this case the origin) to the focus is traditionally labeled as ” p “.

table>

The leading coefficient in y = ax 2 + bx + c is labeled “a “. So when examining the coefficient of x 2, we are examining a, p is the distance from the vertex to the focus.

You remember the vertex form of a parabola as being y = a ( x – h ) 2 + k where ( h, k ) is the vertex of the parabola. If we let the coefficient of x 2 (or a ) = and perform some algebraic maneuvering, we can get the next equation.

 Parabola with Vertex at ( h, k ) (axis of symmetry parallel to the y -axis) (Known as ” standard form “.) Conic Form of Parabola Equation: ( x – h ) 2 = 4 p ( y – k ) with the vertex at ( h, k ), the focus at ( h, k+p ) and the directrix y = k – p Since the example at the right is a translation of the previous graph, the relationship between the parabola and its focus and directrix remains the same ( p = ¼). So with a vertex of (2,-3), we have: ( x – 2) 2 = 4(¼) ( y – (-3)) ( x – 2) 2 = y + 3 The focus is at (2,-3+¼) or (2,-2¾) and the directrix is y = -3-¼ or y = -3¼ This “new” equation is just another form of the old “vertex form” of a parabola. OLD: y = ( x – 2)² – 3 NEW: ( x – 2)² = y + 3

S O M E T H I N G N E W ! ! ! Up to this point, all of your parabolas have been either opening upward or opening downward, depending upon whether the leading coefficient was positive or negative respectively. The axis of symmetry of those parabolas is parallel to the y -axis.

 “Sideways” Parabola Vertex (0,0) (axis of symmetry parallel to x -axis) Conic Forms of Parabola Equations: with the vertex at ( 0,0 ), focus at ( p, 0), and directrix x = -p We will now be examining the coefficient of y ², instead of x ². In the example at the right, the coefficient of y ² is 1, so, making p = ¼. The vertex is (0,0), the focus is (¼,0), and the directrix is x = -¼. For parabolas opening to the right or to the left, the y -variable is being squared (instead of the x² we are used to seeing for parabolas).

table>

“Sideways” Parabolas Vertex ( h,k ) (axis of symmetry parallel to x -axis) Conic Form of Parabola Equation: ( y – k ) 2 = 4 p ( x – h ) with the vertex at ( h, k ), the focus at ( h+p, k ) and the directrix x = h – p Sideways Equation in Standard Vertex Form: x = a ( y – k ) 2 + h with the vertex at ( h, k ). Since the example at the right is a translation of the previous graph, the relationship between the parabola and its focus and directrix remains the same. In the example at the right, ( y – 1) 2 = 4(¼) ( x – (-2)) ( y – 1) 2 = x + 2 The vertex is (-2,1), the focus is (-1¾,1) and the directrix is y = -2¼. Notice that parabolas that open right or left, are NOT functions. They fail the vertical line test for functions. These parabolas are considered relations. The Standard “Vertex Form” and the Conic Form are the same: VERTEX FORM: x = ( y – 1)² – 2 CONIC FORM: ( y – 1)² = x + 2

S u m m i n g U p: ( p is distance from vertex to focus) Deriving the Conic Parabola Equation:

 Deriving the Parabola Equation Start by placing the parabola’s vertex at the origin, for ease of computation. By definition, the distance, p, from the origin to the focus will equal the distance from the origin to the directrix (which will be y = -p ). The focus is point F and FA = AB by definition. Using the Distance Formula, we know Since we know FA = AB, we have The distance from the vertex (origin) to the focus is traditionally labeled as ” p “. (” p ” is also the distance from the vertex to the directrix.)

Given x 2 = 16y, state whether the parabola opens upward, downward, right or left, and state the coordinates of the focus.

 ANSWER: Form: x 2 = 4 py 4 p = 16 p = 4 The focal length is 4. Since this “form” squares x, and the value of 4 p is positive, the parabola opens upward. This form of parabola has its vertex at the origin, (0,0). The focal length (distance from vertex to focus) is 4 units. The focus is located at (0,4).

Given the parabola, (x – 3) 2 = -8(y – 2), state whether the parabola opens upward, downward, right or left, and state the coordinates of the vertex, the focus, and the equation of the directrix.

 ANSWER: Form: ( x – h ) 2 = 4 p ( y – k ) Vertex: ( h,k ) = (3,2) 4 p = 8 p = 2 The focal length is 2. Since this “form” squares x, and the value of 4 p is negative, the parabola opens downward. This form of parabola has its vertex at ( h,k ) = (3,2). The focal length (distance from vertex to focus) is 2 units. The focus is located at (3,0). The directrix is y = 4.

Write the equation of a parabola with a vertex at the origin and a focus of (0,-3).

 ANSWER: Make a sketch. Remember that the parabola opens “around” the focus. Vertex: (0,0) and Focus: (0,-3) Focal length p = 3. Opening downward means negative. Form of Equation: x 2 = 4 py EQUATION: x 2 = 4(-3) y x 2 = -12 y

Find the focus and directrix of the parabola whose equation is x 2 – 6x + 3y + 18 = 0.

 ANSWER: You need to complete the square so the vertex, focus and directrix information will be visible. • The vertex is (3,-3). • The x -squared term indicates the parabola opens upward or downward. • The negative value indicates the parabola opens downward. • The focal length, p, is: 4 p = 3; p = ¾ • The focus is at (3, -3¾) • The directrix is y = -2¼

Write the equation of a parabola whose focus (-2,1) and whose directrix is x = -6.

 ANSWER: Make a sketch. Remember that the parabola opens “around” the focus, and the vertex is halfway between the focus and the directrix. • Vertex: (-4,1) = ( h,k ) • Opens to the right (around the focus) • Focal length, p = 2 • Form of Equation: ( y – k ) 2 = 4 p ( x – h ) EQUATION: ( y – 1) 2 = 4(2)( x – (-4)) ( y – 1) 2 = 8( x + 4)

/td>

#### What is vertex in math?

Vertex is a point on a polygon where the sides or edges of the object meet or where two rays or line segments meet. The plural of a vertex is vertices. For example, in the above figures, points A, B, C, D, and E are vertices.

#### Is the vertex always positive?

What is the vertex of a parabola? – Parabolas always have a lowest point (or a highest point, if the parabola is upside-down). This point, where the parabola changes direction, is called the “vertex”. If the quadratic is written in the form y = a ( x − h ) 2 + k, then the vertex is the point ( h, k ),

1. This makes sense, if you think about it.
2. The squared part of y = a ( x − h ) 2 + k is always positive (for a right-side-up parabola), unless it’s zero.
3. So you’ll always have that fixed value k, and then you’ll always be adding something to it to make y bigger, unless of course the squared part is zero.

So the smallest y can possibly be is y = k, and this smallest value will happen when the squared part, x − h, equals zero. And the squared part is zero when x − h = 0, or when x = h,

The same reasoning works, with k being the largest value and the squared part always subtracting from it, for upside-down parabolas.(Note: The ” a ” in the vertex form ” y = a ( x − h ) 2 + k ” of the quadratic is the same as the ” a ” in the common form of the quadratic equation, ” y = ax 2 + bx + c “.)Since the vertex is a useful point, and since you can “read off” the coordinates for the vertex from the vertex form of the quadratic, you can see where the vertex form of the quadratic can be helpful, especially if the vertex isn’t one of your T-chart values.

However, quadratics are not usually written in vertex form. You can to convert ax 2 + bx + c to vertex form, but, for finding the vertex, it’s simpler to just use, (The vertex formula is derived from the completing-the-square process, just as is the Quadratic Formula.

Find the vertex of y = 3 x 2 + x − 2 and graph the parabola.

To find the vertex, I look at the coefficients a, b, and c, The formula for the vertex gives me: Then I can find k by evaluating y at : So now I know that the vertex is at, Using the formula was helpful, because this point is not one that I was likely to get on my T-chart. I need additional points for my graph: Now I can do my graph, and I will label the vertex: When you write down the vertex in your homework, write down the exact coordinates: ” “. But for graphing purposes, the decimal approximation of ” (−0.2, −2.1) ” may be more helpful, since it’s easier to locate on the axes. The only other consideration regarding the vertex is the “axis of symmetry”.

#### Why is H negative in vertex form?

The parameter h is the extremal point of the quadratic. The extremum happens when x=h, when the function (x−h)2 is minimized (or −(x−h)2 maximized). So the reason for the minus sign is because x=h is equivalent to x−h=0.

#### What is the vertex of a triangle?

The point of intersection of any two sides of a triangle is known as a vertex. A triangle has three vertices.

#### What is the vertex in math?

Algebra Applied Mathematics Calculus and Analysis Discrete Mathematics Foundations of Mathematics Geometry History and Terminology Number Theory Probability and Statistics Recreational Mathematics Topology Alphabetical Index New in MathWorld A vertex is a special point of a mathematical object, and is usually a location where two or more lines or edges meet.

#### Which is the vertex?

Of an angle – A vertex of an angle is the endpoint where two lines or rays come together. The vertex of an angle is the point where two rays begin or meet, where two line segments join or meet, where two lines intersect (cross), or any appropriate combination of rays, segments, and lines that result in two straight “sides” meeting at one place.