a) Find the vertex, standard graphing form, and xintercepts for $f(x)={\frac {1}{3}}x^{2}+2x3$
b) Sketch the graph. Provide the yintercept.
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?

Answer:

1) Standard graphing form is $yh=a(xk)^{2}$.

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 $y={\frac {1}{3}}x^{2}+2x3$ by 3, we get

$3y=x^{2}+6x9$.

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 $(33{\sqrt {2}},0)$.


Step 4:

To find the $y$ intercept, we let $x=0$. So, we get $y=3$.

Thus, the $y$ intercept is (0,3).


Final Answer:

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 $(33{\sqrt {2}},0)$.

The $y$ intercept is (0,3)

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