Question

The graph of y=ax^2+bx+cx has a minimum at (5, - 3) and passes through (4, 0). Find the values of a, b, and c​

Answer 1

Y=ax^2+bx+CX=(5,-3)(4,0)

Answer 2

Write the equation in vertex form
Y = a(x- j)^2 + k. Where j, k. Is the coordinates of the vertex
So we have
Y = a(x - 5)^2 - 3
Now we find the value of a by substituting the point (4, 0):
0 = a(4-5)^2 - 3
—>. a* (-1)^2 - 3 = 0
—-> a = 3.

3(x - 5)^2 - 3 = y
Y = 3(x^2 - 10x + 25) - 3
Y = 3x^2 - 30x + 72
So a = 3, b = -30 and c = 72