find the following for the function
f(x)= (x+5)^2(x-1)^2
a) find the x and y intercepts for the polynomial function f
b) determine whether the graph of f crosses or touches the x axis at each x intercept
C) put all information together to obtain the graph of f
The x-intercepts are the roots.
The y-intercept is at 5^2*1^2 = 25
Since the roots are of even order, the graph touches the axis at each root.
So, degree-4 polynomial with positive coefficient. Down from the left, tangent at each root, then back up to the right.
To find the x and y intercepts of the polynomial function f(x) = (x+5)^2(x-1)^2, we need to set the function equal to zero and find the values of x that satisfy this equation.
a) X-intercepts:
Setting f(x) = 0, we have
(x+5)^2(x-1)^2 = 0
To solve this equation, we set each factor equal to zero:
x + 5 = 0, which gives x = -5
x - 1 = 0, which gives x = 1
Therefore, the x-intercepts of the function f(x) are x = -5 and x = 1.
b) Y-intercept:
To find the y-intercept, we substitute x = 0 into the function:
f(0) = (0+5)^2(0-1)^2
= 5^2 * (-1)^2
= 25 * 1
= 25
Thus, the y-intercept of the function f(x) is y = 25.
c) Graph of f:
To plot the graph of f(x), we consider the x-intercepts and the behavior of the function near each intercept.
Since x = -5 is a repeated root (double root), the graph touches the x-axis at x = -5.
At x = 1, the x-intercept is also a repeated root, so the graph touches the x-axis at x = 1 as well.
The y-intercept is at (0,25).
Based on this information, we can conclude that the graph of f(x) touches the x-axis at x = -5 and x = 1, and the graph is above the x-axis for all other values of x.
To sketch the graph, plot the x-intercepts at x = -5 and x = 1, and indicate that the graph touches (but does not cross) the x-axis at these points. Finally, connect the points smoothly, with the graph remaining above the x-axis for all other values of x.