Could someone help me with finding the vertex from an equation?

[TrulySusan]

I can usually solve these, but this has me stuck:

 

y= -2x^2 + 8x -3 ( negative 2x squared plus 8x minus 3)

 

Thanks for any help you can give- please show me the steps if possible.

[regentrude]

Do you have calculus available?

Find maximum by taking derivative dy/dx and setting it to zero, find x. Put back into equation, find y

-4x+8=0, x=2

y(x=2)=5

Edited December 12, 2018 by regentrude

[regentrude]

if no calculus:

parabola written as a(x-h)^2+k has vertex (h,k)

so, ax^2-2ahx+ah^2+k = -2x^2+8x-3

Comparing coefficients:

square term: a=- 2

linear term: -2ah=8

-2(-2)h= 8

h=2

constant term: ah^2+k=-3

(-2)*2^2+k=-3

k=5

Vertex is at (2,5)

[TrulySusan]

Thanks so much!! This is for a public school algebra 1 class study sheet to prepare for the final. No answers were provided( not helpful for studying) and ds is pretty sure they didn't cover this in class.  And they have no textbooks.

 

[chocolate-chip chooky]

If you have your quadratic in the form of y = ax^2 + bx + c, then  -b/2a should give you the x coordinate of your turning point.

In your equation, b= 8 and a= -2, so -b/2a = -8/-4 = 2

so 2 is your x-coordinate of your turning point.

Then plug that back into your equation to find the y coordinate: y = 5

So your turning point is (2,5).

@regentrude is this sound?

[regentrude]

2 hours ago, chocolate-chip chooky said:

If you have your quadratic in the form of y = ax^2 + bx + c, then  -b/2a should give you the x coordinate of your turning point.

In your equation, b= 8 and a= -2, so -b/2a = -8/-4 = 2

so 2 is your x-coordinate of your turning point.

Then plug that back into your equation to find the y coordinate: y = 5

So your turning point is (2,5).

@regentrude is this sound?

Yes, that's correct.