Vertex Form
To easily graph a quadratic equation, we can convert it to vertex form.
This post is a chapter in the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Vertex Form. Justin Math: Algebra. https://justinmath.com/vertex-form/
To easily graph a quadratic equation, we can convert it to vertex form:
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).
Converting to Vertex Form
To convert a quadratic equation into vertex form, we can complete the square.
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 a chapter in the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Vertex Form. Justin Math: Algebra. https://justinmath.com/vertex-form/