find the x-intercepts and y-intercepts
f(x)=-x^2+4x+32
F(x) - Y = -X ^2 + 4X + 32
-X^2 + 4X + 32 = 0
Use quadratic formula:
X = (- 4 +- sqrt(16 + 128))/-2
X = (-4 +- 12) /-2 = -4, and 8 =
X - Intercepts.
Let X = 0
Y =-(0)^2 + 4(0) + 32 = 32 = Y-Int.
To find the x-intercepts of a function, you need to set f(x) equal to zero and solve for x. In other words, you are looking for the values of x at which the function crosses or touches the x-axis.
Let's find the x-intercepts of the given function f(x) = -x^2 + 4x + 32:
Step 1: Set f(x) = 0:
- x^2 + 4x + 32 = 0
Step 2: Solve the quadratic equation. You can use factoring, completing the square, or the quadratic formula. In this case, it's not easy to factor, so we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -1, b = 4, and c = 32. Plugging these values into the quadratic formula, we get:
x = (-4 ± √(4^2 - 4(-1)(32))) / (2(-1))
x = (-4 ± √(16 + 128)) / (-2)
x = (-4 ± √144) / (-2)
x = (-4 ± 12) / (-2)
Simplifying further, we have two possible x-values:
x1 = (-4 + 12) / (-2) = 8 / (-2) = -4
x2 = (-4 - 12) / (-2) = -16 / (-2) = 8
So the x-intercepts are x = -4 and x = 8.
To find the y-intercept, we need to evaluate f(x) when x is equal to zero. The y-intercept is the point where the graph crosses or touches the y-axis.
Step 1: Set x = 0 in the function f(x):
f(0) = -(0)^2 + 4(0) + 32
f(0) = 0 + 0 + 32
f(0) = 32
So the y-intercept is (0, 32).