H(x)=-x^2+10x-16 What is the height of the ball at the time it is thrown

0m

Step-by-step explanation:

Before the ball is thrown, it is at the ground level. At the ground level, the height is zero. Substitute h = 0 into the equation;

h(x)=-x^2+10x-16

0=-x^2+10x-16

x^2-10x+16 = 0

Factorize

x^2-8x-2x+16 = 0

x(x-8) - 2 (x-8) = 0

x-2 = 0 and x - 8 = 0

x = 2 and 8

At x = 2

h(2) = -2^2 + 10(2) - 16

h(2) = -4 + 4

h(2) = 0

Hence the height of the ball at the time it is thrown is 0m