# 004 Sample Final A, Problem 2

a) Find the vertex, standard graphing form, and x-intercepts for ${\displaystyle f(x)={\frac {1}{3}}x^{2}+2x-3}$
b) Sketch the graph. Provide the y-intercept.

Foundations
1) What is the standard graphing form of a parabola?
2) What is the vertex of a parabola?
3) What is the ${\displaystyle y}$-intercept?
1) Standard graphing form is ${\displaystyle y-h=a(x-k)^{2}}$.
2) Using the standard graphing form, the vertex is ${\displaystyle (h,k)}$.
3) The ${\displaystyle y}$-intercept is the point ${\displaystyle (0,y)}$ where ${\displaystyle f(0)=y}$.

Solution:

Step 1:
First, we put the equation into standard graphing form. Multiplying the equation ${\displaystyle y={\frac {1}{3}}x^{2}+2x-3}$ by 3, we get
${\displaystyle 3y=x^{2}+6x-9}$.
Step 2:
Completing the square, we get ${\displaystyle 3y=(x+3)^{2}-18}$. Dividing by 3 and subtracting 6 on both sides, we have
${\displaystyle y+6={\frac {1}{3}}(x+3)^{2}}$.
Step 3:
From standard graphing form, we see that the vertex is (-3,-6). Also, to find the ${\displaystyle x}$ intercept, we let ${\displaystyle y=0}$. So,
${\displaystyle 18=(x+3)^{2}}$. Solving, we get ${\displaystyle x=-3\pm 3{\sqrt {2}}}$.
Thus, the two ${\displaystyle x}$ intercepts occur at ${\displaystyle (-3+3{\sqrt {2}},0)}$ and ${\displaystyle (-3-3{\sqrt {2}},0)}$.
Step 4:
To find the ${\displaystyle y}$ intercept, we let ${\displaystyle x=0}$. So, we get ${\displaystyle y=-3}$.
Thus, the ${\displaystyle y}$ intercept is (0,-3).
The vertex is (-3,-6). The equation in standard graphing form is ${\displaystyle y+6={\frac {1}{3}}(x+3)^{2}}$.
The two ${\displaystyle x}$ intercepts are ${\displaystyle (-3+3{\sqrt {2}},0)}$ and ${\displaystyle (-3-3{\sqrt {2}},0)}$.
The ${\displaystyle y}$ intercept is (0,-3)