Let's go through each part of the question:
A) The value of f(-5):
To find the value of f(-5), we substitute x = -5 into the equation f(x) = x^2 + 6x - 8:
f(-5) = (-5)^2 + 6(-5) - 8
f(-5) = 25 - 30 - 8
f(-5) = -13
B) The vertex:
To find the vertex of the parabola, we use the formula: x = -b/(2a), where a and b are the coefficients of the quadratic equation.
For the equation f(x) = x^2 + 6x - 8, a = 1 and b = 6.
Using the formula, we find:
x = -6/(2*1) = -6/2 = -3
To find the corresponding y-coordinate, we substitute x = -3 into the equation:
f(-3) = (-3)^2 + 6(-3) - 8
f(-3) = 9 - 18 - 8
f(-3) = -17
So the vertex is (-3, -17).
C) The y-intercept:
To find the y-intercept, we substitute x = 0 into the equation:
f(0) = (0)^2 + 6(0) - 8
f(0) = 0 - 0 - 8
f(0) = -8
So the y-intercept is (0, -8).
D) Find the values of x that make f(x) = 0:
To find the values of x that make f(x) = 0, we set the equation equal to zero and solve for x:
x^2 + 6x - 8 = 0
Using the quadratic formula or factoring, we find the solutions:
x = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values a = 1, b = 6, and c = -8, we get:
x = (-6 ± √(6^2 - 4(1)(-8)))/(2(1))
x = (-6 ± √(36 + 32))/2
x = (-6 ± √68)/2
x = (-6 ± 2√17)/2
x = -3 ± √17
So the two values of x that make f(x) = 0 are approximately -3 + √17 and -3 - √17.