Given y=ax^2-10x+c the vertex is (-5,20). Find the missing coefficients

Answer 1

If y = ax² + bx + c and a is nonzero, then the x-coordinate of the vertex is -b/(2a)

So, -b/(2a) = -5 and b = -10. Therefore, 10 = -10a. So, a = -1.

y = -x² - 10x + c

Since (-5,20) is on the graph, 20 = -25 + 50 + c. So, c = -5.

y = -x² - 10x - 5

Answer 2

Given the vertex is located at (-5,20) the equation of the parabola can be written as:

y = a(x + 5)² + 20 [vertex form; there are infinitely many parabolas with that vertex, depending on a.

We also know: y = ax² -10x + c

Therefore:

a(x + 5)² + 20 = ax² -10x + c

a(x² + 10x + 25) + 20 = ax² - 10x + c

ax² +10ax +25a + 20 = ax² -10x + c

10ax + 25a + 20 = -10x + c

Here is where you can equate coefficients to get two equations:

10a = -10 [the coefficients of x on each side must be equal]

25a + 20 = c [constants must be equal]

a = -1

-25 + 20 = c

c = -5

So,

y = -x² - 10x - 5