Can you please check my work and answers.
Find the x-intercepts of the polynomial function. State whether the graph crosses the x-axis, or touches the x-axis and turns around, at each intercept.
1.f(x) = (x + 1)(x - 3)(x - 1)^2
-I got -1 crosses the x-axis, 3 crosses the x-axis, 1 touches the x-axis and turns around
5.f(x) = x^4 - 16x^2
- I got x^2(x^2-16)
(x+4)(x-4)
0 touches the x-axis and turns around; 4 crosses the x-axis; -4 crosses the x-axis
7.x^4 + 5x^3 - 84x^2 = 0
I got 0 touches the x-axis and turns around;-12 crosses the x-axis; 7 crosses the x-axis
To find the x-intercepts of a polynomial function, you need to set the function equal to zero and solve for x. The x-intercepts are the points where the graph of the function intersects the x-axis.
Let's go through each of your problems:
1. f(x) = (x + 1)(x - 3)(x - 1)^2
To find the x-intercepts, set f(x) = 0:
0 = (x + 1)(x - 3)(x - 1)^2
So, x + 1 = 0, x - 3 = 0, and (x - 1)^2 = 0.
Solving these equations, we get x = -1, x = 3, and x = 1.
You correctly identified the x-intercepts as -1, 3, and 1.
To determine whether the graph crosses or touches the x-axis, we look at the factors.
At x = -1 and x = 3, the factors are linear, so the graph crosses the x-axis at these points.
At x = 1, the factor (x - 1)^2 is squared, indicating that the graph touches the x-axis and turns around.
5. f(x) = x^4 - 16x^2
Factorizing the equation:
f(x) = x^2(x^2 - 16)
f(x) = x^2(x + 4)(x - 4)
Setting f(x) = 0:
x^2 = 0, x + 4 = 0, and x - 4 = 0
Solving these equations, we get x = 0, x = -4, and x = 4.
You correctly identified the x-intercepts as 0, -4, and 4.
At x = 0, the factor x^2 is squared, indicating that the graph touches the x-axis and turns around.
At x = -4 and x = 4, the factors are linear, so the graph crosses the x-axis at these points.
7. f(x) = x^4 + 5x^3 - 84x^2 = 0
To factorize this equation, we look for common factors.
f(x) = x^2(x^2 + 5x - 84)
f(x) = x^2(x + 12)(x - 7)
Setting f(x) = 0:
x^2 = 0, x + 12 = 0, and x - 7 = 0
Solving these equations, we get x = 0, x = -12, and x = 7.
You correctly identified the x-intercepts as 0, -12, and 7.
At x = 0, the factor x^2 is squared, indicating that the graph touches the x-axis and turns around.
At x = -12 and x = 7, the factors are linear, so the graph crosses the x-axis at these points.
In summary:
1. f(x) = (x + 1)(x - 3)(x - 1)^2
-1 crosses the x-axis, 3 crosses the x-axis, 1 touches the x-axis and turns around.
5. f(x) = x^4 - 16x^2
0 touches the x-axis and turns around; -4 crosses the x-axis; 4 crosses the x-axis.
7. f(x) = x^4 + 5x^3 - 84x^2
0 touches the x-axis and turns around; -12 crosses the x-axis; 7 crosses the x-axis.