# 004 Sample Final A, Problem 2

a) Find the vertex, standard graphing form, and x-intercepts for $f(x)={\frac {1}{3}}x^{2}+2x-3$ 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 $y$ -intercept?
1) Standard graphing form is $y-h=a(x-k)^{2}$ .
2) Using the standard graphing form, the vertex is $(h,k)$ .
3) The $y$ -intercept is the point $(0,y)$ where $f(0)=y$ .

Solution:

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