Vertex Form

by Justin Skycak (x.com/justinskycak) on

To easily graph a quadratic equation, we can convert it to vertex form.

This post is part of the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Vertex Form. In Justin Math: Algebra. https://justinmath.com/vertex-form/

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To easily graph a quadratic equation, we can convert it to vertex form:

$y=a(x-h)^2+k$


In vertex form, we can tell the coordinates of the vertex of the parabola just by looking at the equation: the vertex is at $(h,k)$. We can also tell which way the parabola opens, by checking whether $a$ is positive (opens up) or negative (opens down).

$\begin{align*} \textbf{Given Equation} \hspace{.5cm} & \hspace{.5cm} \textbf{Vertex} \hspace{.5cm} && \hspace{.5cm} \textbf{Opens} \\ y=(x-2)^2+1 \hspace{.5cm} &\Bigg| \hspace{.5cm} (2,1) \hspace{.5cm} &&\Bigg| \hspace{.5cm} \text{up} \\ y=-2(x-5)^2-3 \hspace{.5cm} &\Bigg| \hspace{.5cm} (5,-3) \hspace{.5cm} &&\Bigg| \hspace{.5cm} \text{down} \\ y=7(x-1)^2+\frac{1}{2} \hspace{.5cm} &\Bigg| \hspace{.5cm} \left( 1, \frac{1}{2} \right) \hspace{.5cm} &&\Bigg| \hspace{.5cm} \text{up} \\ y=-\frac{1}{3} \left( x+\frac{2}{7} \right)^2-\frac{4}{5} \hspace{.5cm} &\Bigg| \hspace{.5cm} \left( -\frac{2}{7}, -\frac{4}{5} \right) \hspace{.5cm} &&\Bigg| \hspace{.5cm} \text{down} \end{align*}$


Converting to Vertex Form

To convert a quadratic equation into vertex form, we can complete the square.

$\begin{align*} \text{Original equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} y=2x^2-4x-5 \\ \text{Divide by } 2 \hspace{.5cm} &\Bigg| \hspace{.5cm} \frac{y}{2}=x^2-2x-\frac{5}{2} \\ \begin{matrix} \text{Move the constant to} \\ \text{the other side} \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} \frac{y}{2}+\frac{5}{2}=x^2-2x \\ \text{Add } \left( \frac{-2}{2} \right)^2 = 1 \text{ to both sides} \hspace{.5cm} &\Bigg| \hspace{.5cm} \frac{y}{2} + \frac{7}{2} =x^2-2x+1 \\ \text{Factor} \hspace{.5cm} &\Bigg| \hspace{.5cm} \frac{y}{2} + \frac{7}{2} =(x+1)^2 \\ \text{Multiply by } 2 \hspace{.5cm} &\Bigg| \hspace{.5cm} y+7 =2(x+1)^2 \\ \text{Subtract } 7 \hspace{.5cm} &\Bigg| \hspace{.5cm} y =2(x+1)^2-7 \\ \end{align*}$


Exercises

Write the equation in vertex form $y=a(x-h)^2+k$. Then, find the coordinates of the vertex and tell which way the parabola opens. (You can view the solution by clicking on the problem.)

$1) \hspace{.5cm} y=x^2+2x+3$
Solution:
$y=(x+1)^2+2$
$(-1,2) \hspace{.5cm} \text{up}$

$2) \hspace{.5cm} y=x^2-6x+4$
Solution:
$y=(x+1)^2+2$
$(3,-5) \hspace{.5cm} \text{up}$

$3) \hspace{.5cm} y=2x^2+20x-5$
Solution:
$y=2(x+5)^2-55$
$(-5,-55) \hspace{.5cm} \text{up}$

$4) \hspace{.5cm} y=-3x^2+6x+1$
Solution:
$y=-3(x-1)^2+4$
$(1,4) \hspace{.5cm} \text{down}$

$5) \hspace{.5cm} y=2x^2-2x-1$
Solution:
$y=2 \left( x-\frac{1}{2} \right)^2-\frac{3}{2}$
$\left( \frac{1}{2}, -\frac{3}{2} \right) \hspace{.5cm} \text{up}$

$6) \hspace{.5cm} y=-\frac{1}{3}x^2-2x-10$
Solution:
$y=-\frac{1}{3} (x+3)^2 - 7$
$(-3,-7) \hspace{.5cm} \text{down}$

$7) \hspace{.5cm} y=\frac{3}{4}x^2-2x+3$
Solution:
$y=\frac{3}{4} \left( x-\frac{4}{3} \right)^2+\frac{5}{3}$
$\left( \frac{4}{3}, \frac{5}{3} \right) \hspace{.5cm} \text{up}$

$8) \hspace{.5cm} y=-\frac{2}{3}x^2-8x-17$
Solution:
$y=-\frac{2}{3} (x+6)^2 + 7$
$(-6,7) \hspace{.5cm} \text{down}$

This post is part of the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Vertex Form. In Justin Math: Algebra. https://justinmath.com/vertex-form/


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