# Vertex Form of a Quadratic Equation

## Key Questions

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#### Explanation:

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Quadratic Equations in Vertex Form have a general form:

#color(red)(y=f(x)=a(x-h)^2+k#, where

#color(red)((h,k)# is the #color(blue)("Vertex"#

Let us consider a quadratic equation in Vertex Form:

#color(blue)(y=f(x)=(x-3)^2+8#, where

#color(green)(a=1; h=3; k=8#

Hence, #color(blue)("Vertex "= (3, 8)#

To find the y-intercept, set #color(red)(x=0#

$y = {\left(0 - 3\right)}^{2} + 8$

$y = 9 + 8$

$y = 17$

Hence, the y-intercept: #color(blue)((0, 17)#

We can use a table of values to draw the graph:

Use the table with two columns #color(red)(x and y# to draw the graph as shown below:

The Parent Graph of #color(red)(y=x^2# can also be seen for comparison, to better understand transformation.

Also note that,

Axis of Symmetry is #color(red)(x=h#

$\Rightarrow x = 3$

We can verify this from the graph below:

Hope it helps.

• Since the equation is:

$y = {x}^{2} + b x + c$

the vertex is $V \left(- \frac{b}{2 a} , - \frac{\Delta}{4 a}\right)$,

or, found the ${x}_{v} = - \frac{b}{2 a}$ you can substitue it in the equation of the parabola at the place of $x$, finding the ${y}_{v}$.

• Vertex Form

$y = a {\left(x - h\right)}^{2} + k$,

where $\left(h , k\right)$ is the vertex.

I hope that this was helpful.