Question

In the xy-coordinate plane, the graph of the equation y = 2x^2 - 12x-32 has zeros at x = d and x = e, where d > e. The graph has a minimum at (f,-50). What are the values of d, e, and ƒ?

A. d = 2, e = 8, and f = -1/8

B. d = 8, e = -2, and ƒ = 3

C. d = -2, e = 8, and ƒ = 2

D. d = 2, e = 8, and ƒ = -3

Answer 1

To find the values of d, e, and f, we first need to determine the zeros of the equation y = 2x^2 - 12x - 32. The zeros occur when y = 0, so we set the equation equal to 0:

2x^2 - 12x - 32 = 0

Factoring the equation, we get:

2(x^2 - 6x - 16) = 0
2(x - 8)(x + 2) = 0

Setting each factor to 0 gives us the zeros:

x - 8 = 0, x = 8
x + 2 = 0, x = -2

Therefore, d = 8 and e = -2.

To find the x-coordinate of the minimum, we use the formula for the x-coordinate of the vertex of a parabola: x = -b/2a, where a = 2 and b = -12 in this case:

x = -(12)/(2*2) = -3

So, the x-coordinate of the minimum is -3. Substituting this value into the equation gives us:

y = 2*(-3)^2 - 12*(-3) - 32 = 18 + 36 - 32 = 22

Therefore, the values of d, e, and f are 8, -2, and -3 respectively. So, the answer is D. d = 2, e = 8, and f = -3.

Answer 2

The function h(x) = x^2 + 3 and g(x) = x^2 – 6. If g(x) = h(x) + k, what is the value of k?

k=

Answer 3

We are given h(x) = x^2 + 3 and g(x) = x^2 - 6.

To find the value of k such that g(x) = h(x) + k, we need to equate the two functions and solve for k:

x^2 - 6 = x^2 + 3 + k

Subtract x^2 from both sides:

-6 = 3 + k

Subtract 3 from both sides to solve for k:

k = -6 - 3
k = -9

Therefore, the value of k is -9.