Find b and c so that y =2x power 2 +bx+c has a vertex (-9,5)

Question

Find b and c so that y =2x power 2 +bx+c has a vertex (-9,5)

Answer 1

y = 2x² + bx + c

For y = ax² + bx + c, the x-coordinate of the vertex is -b / (2a).

So, -b / 4 = -9. Therefore, b = 36.

y = 2x² + 36x + c

Since (-9,5) is on the graph, 5 = 162 - 324 + c

c = 167

Answer 2

Using vertex form:

y = 2(x + 9)² + 5

Can you expland and answer?

Answer 3

y=2x^2 +bx+ c with vertex (-9,5)

y= 2(x^2 +bx/2 + b^2/4) + c - b^2/2

y = 2(x+b/4)^2 +c-b^2/2

-b/4 =-9

b=36

c-b^2/8 = 5

c=5+b^2/8 =5+36^2/8= 167

b=36, c=167

check the answers

y= 2x^2 +bx + c

5= 2(-9)^2+(36)(-9)+167

5 = 162-324 +167 = 5

b= 36, c=167

y=2x^2 +36x +167

or another approach

y = 2(x- -9)^2 +5 in vertex form

= 2x^2 +36x + 162 + 5

= 2x^2 +36x +167

b=36,

c=167

or another approach

x coordinate of the vertex =-b/2a (look at the quadratic formula, x=-b/2a +/-(1/2a)(sqr(b^2-4ac))

-9=-b/2a

9=b/2(2)

b =9(2)(2) =36

5=2(-9)^2 +36(-9) +c

c=5-2(81)+36(9)

c = -162+324+5

c = 167

It's an upward opening parabola with vertex =minimum point = (-9,5)

y intercept = 167 or the point (0,167)

axis of symmetry is x=-9

another point, due to the symmetry, is (-18,167)

no x intercepts, it's all above the x-axis

it might help to draw a rough sketch of the parabola, plot the points