Please help me solve

Question

find b and c so that y=5x^2+bx+c has vertex(7,-6)

Answer 1

Here is one way to solve this problem:

1. The formula for the axis of symmetry is x = -b/2a, In this case a=5, so x= -b/10. Since the vertex lies on the axis of symmetry, solve -b/10 = 7 to get b.
2. The vertex lies on the parabola, so when x = 7, y = -6. Substitute those values into y = 5x²-70x+c to solve for c.

Check your work here:

desmos.com/calculator/eupky6fqyd

Answer 2

To find the vertex in a quadratic, we can use x = -b/(2a) and since we know the x-coordinate of the vertex is 7 and that a = 5 we can solve for b:

7 = -b/(2•5)

7 = -b/10

-b = 70

b = -70

This gives us f(x) = 5x² - 70x + c

We also know that f(7) = -6 (because the vertex is (7, -6) so we can plug in x = 7 and f(7) = -6 to solve for "c":

-6 = 5(7)² - 70(7) + c

-6 = 5(49) - 490 + c

-6 = 245 - 490 + c

-6 = -245 + c

c = 239