SOLUTION: Determine the x-intercepts of P(x)=(2x-7)^2(x-1)^3 and state whether the graph of P crosses the x-axis or intersects but does not cross the x-axis

Question 376783: Determine the x-intercepts of P(x)=(2x-7)^2(x-1)^3 and state whether the graph of P crosses the x-axis or intersects but does not cross the x-axis
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
X-intercepts are points where a graph intersects the x-axis. All points on the x-axis have a y-coordinate that is zero. So the x-intercepts will occur when P(x) = 0:

We have a product that is zero and the Zero Product Property tells us that this (or any) product can be zero only is one (or more) of the factors is zero. So:
2x-7 = 0 or x-1 = 0
Solving these we get:
x = 7/2 or x = 1
So the x-intercepts are (7/2, 0) and (1. 0).

As to whether the graph of P crosses the x-axis...
Above the x-axis is where y-coordinates are positive and below the x-axis is where y-coordinates are negative. So if the graph P crosses the x-axis

It will cross at an x-intercept, and
The y-coordinate (or value of P) will change from positive to negative (or vice versa) near an x-intercept.

So we need to check the values of P near the x-intercepts.

For x values near 7/2 (aka 3.5), like 3.6 or 3.4, the factor will be positive and the factor will also be positive or zero (because it is being squared). So near the x-intercept (7/2, 0) the value of P is never negative. This means that the graph of P is always at or above the x-axis near 7/2. So P intersects but does not cross the x-axis at 7/2.

For x values near 1, like 1.1 or 0.9, the factor will be positive if x > 1 and it will be negative when x < 1. And the factor will be positive (because it is being squared). So near the x-intercept (1, 0) the value of P is positive for x > 1 and negative for x < 1. This means that the graph of P crosses the x-axis near 1.

FWIW, here's the graph of P(x):

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