find the following for the function f(x)=(x+5)^2(x-2)^2
a.find the x and y intercepts,
b.find the power function that the graph ressembles for large values of x c.determine the maximum number of turning points on the graph of f
d.determine the behavior of the graph of f near each x-intercept
I don't even know where to begin. please help.
surely the intercepts are easy
f(x) is a polynomial of degree 4, so for large x it just looks like x^4.
Since the graph just touches the x-axis at x=-5 and 2, It comes down from the upper left, touches at x=-5, goes back up and comes back down to touch at x=2, then heads on up to the right, it has 3 turning points.
Visit wolframalpha.com and enter
(x+5)^2(x-2)^2
to see the graph. It will help clarify the concepts, I think.
Enter
plot y=(x+5)^2(x-2)^2 and y=x^4 where -10<x<10
to see how the two graphs are similar for large x.
To find the information you're looking for, let's break it down step by step:
a. To find the x-intercepts, we need to find the values of x when f(x) equals zero. In other words, we want to find the values of x that make the function equal to zero. This can be done by setting (x+5)^2(x-2)^2 = 0 and solving for x. To do this, we can use the zero product property, which states that if a*b = 0, then either a = 0 or b = 0. Applying this property to the given function, we get two cases:
(x+5)^2 = 0 or (x-2)^2 = 0
Taking the square root of both sides, we find:
x + 5 = 0 or x - 2 = 0
Solving for x, we get:
x = -5 or x = 2
So the x-intercepts of the function f(x) are x = -5 and x = 2.
To find the y-intercept, we can simply substitute x = 0 into the function and solve for f(x). Thus, we have:
f(0) = (0+5)^2(0-2)^2
f(0) = 5^2 * (-2)^2
f(0) = 25 * 4
f(0) = 100
Therefore, the y-intercept of the function f(x) is y = 100.
b. To determine the power function that the graph resembles for large values of x, we need to consider the highest exponent in the function. In this case, the highest exponent is 2. Hence, for large values of x, the function f(x) behaves like a quadratic function, which is a power function of degree 2.
c. To determine the maximum number of turning points on the graph of f, we need to consider the power of the function as well. Since the function f(x) has the form (x+5)^2(x-2)^2, it is a quartic function (power function of degree 4). A quartic function can have a maximum of 3 turning points. Therefore, the graph of f can have at most 3 turning points.
d. To determine the behavior of the graph of f near each x-intercept, we need to analyze the sign of f(x) for values of x close to -5 and 2. We can do this by selecting x-values slightly greater and slightly lesser than -5 and 2.
For x close to -5, values like x = -6 and x = -4 can be used. Evaluating f(x) at these values:
f(-6) = (-6+5)^2(-6-2)^2
f(-6) = (-1)^2 * (-8)^2
f(-6) = 1 * 64
f(-6) = 64
f(-4) = (-4+5)^2(-4-2)^2
f(-4) = (1)^2 * (-6)^2
f(-4) = 1 * 36
f(-4) = 36
We observe that as x approaches -5, f(x) becomes positive. Similarly, when x approaches 2, f(x) is also positive. Therefore, the graph of f increases near both x-intercepts.