Question

Rewrite the equation in vertex form. Name the vertex and y intercept:

Answer 1

$y=(3/5)x^2+30x+382$

The vertex form is

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

To get vertex form , we apply completing the square method

For completing the square method we need x^2 alone

factor out 3/5 from first two terms

$y= \frac{3}{5} x^2+30x+382$

When we factor out 3/5, we multiply by 5/3

$y= \frac{3}{5}(x^2+50x)+ 382$

Now we take middle term divide by 2  and square it

50 divide by 2 = 25

25^2 = 625

Now we add and subtract 625

$y= \frac{3}{5}(x^2+50x+625 -625)+ 382$

Take out -625 and multiply with 3/5  and add it with 382

$y= \frac{3}{5}(x^2+50x+625) -625* \frac{3}{5}+ 382$

$y= \frac{3}{5}(x^2+50x+625) -375+ 382$

$y= \frac{3}{5}(x^2+50x+625) +7$

Now factor x^2 + 50x+625 and write it in square form

$y= \frac{3}{5}(x+25)(x+25) +7$

$y= \frac{3}{5}(x+25)^2 +7$

We got vertex form

Vertex is (h,k)

From our equation , h=-25  and k =7

So vertex is (-25,7)

To find y intercept we plug in 0 for x in the original equation

$y= \frac{3}{5}(0)^2+30(0)+382$

y=382

So y intercept is (0,382)