Question

describe the behavior of the graph at the x-intercepts for the function $f(x)=(2x - 7)^7(x + 3)^4$. be sure to identify each x-intercept and justify your answer.

Explanation:

Step1: Find x - intercepts

To find the x - intercepts, we set \(f(x)=0\). So, \((2x - 7)^{7}(x + 3)^{4}=0\).
Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
For \((2x - 7)^{7}=0\), we solve \(2x-7 = 0\), which gives \(2x=7\) or \(x=\frac{7}{2}\).
For \((x + 3)^{4}=0\), we solve \(x + 3=0\), which gives \(x=-3\).

Step2: Analyze the behavior at each x - intercept

For \(x=\frac{7}{2}\):

The exponent of the factor \((2x - 7)\) is \(7\), which is an odd number.
When the exponent of a factor \((x - a)\) in the factored form of a polynomial is odd, the graph of the function crosses the x - axis at \(x = a\).
As \(x\) approaches \(\frac{7}{2}\) from the left, say \(x=\frac{7}{2}-\epsilon\) (where \(\epsilon>0\) is a small number), \(2x - 7=2(\frac{7}{2}-\epsilon)-7=7 - 2\epsilon-7=-2\epsilon<0\), and \((2x - 7)^{7}<0\). The factor \((x + 3)^{4}\) is always non - negative (since any real number raised to an even power is non - negative), so \(f(x)=(2x - 7)^{7}(x + 3)^{4}<0\) when \(x\) is just left of \(\frac{7}{2}\).
As \(x\) approaches \(\frac{7}{2}\) from the right, say \(x=\frac{7}{2}+\epsilon\) (where \(\epsilon>0\) is a small number), \(2x - 7=2(\frac{7}{2}+\epsilon)-7=7 + 2\epsilon-7 = 2\epsilon>0\), and \((2x - 7)^{7}>0\). The factor \((x + 3)^{4}\) is still non - negative, so \(f(x)=(2x - 7)^{7}(x + 3)^{4}>0\) when \(x\) is just right of \(\frac{7}{2}\). So the graph crosses the x - axis at \(x=\frac{7}{2}\).

For \(x=-3\):

The exponent of the factor \((x + 3)\) is \(4\), which is an even number.
When the exponent of a factor \((x - a)\) in the factored form of a polynomial is even, the graph of the function touches the x - axis (and turns around) at \(x = a\).
As \(x\) approaches \(-3\) from the left, say \(x=-3-\epsilon\) (where \(\epsilon>0\) is a small number), \(x + 3=-3-\epsilon + 3=-\epsilon<0\), and \((x + 3)^{4}=(-\epsilon)^{4}=\epsilon^{4}>0\). The factor \((2x - 7)^{7}\) at \(x=-3-\epsilon\) is \((2(-3-\epsilon)-7)^{7}=(-6-2\epsilon - 7)^{7}=(-13 - 2\epsilon)^{7}<0\), so \(f(x)=(2x - 7)^{7}(x + 3)^{4}<0\) (negative times positive is negative).
As \(x\) approaches \(-3\) from the right, say \(x=-3+\epsilon\) (where \(\epsilon>0\) is a small number), \(x + 3=-3+\epsilon+3=\epsilon>0\), and \((x + 3)^{4}=\epsilon^{4}>0\). The factor \((2x - 7)^{7}\) at \(x=-3+\epsilon\) is \((2(-3+\epsilon)-7)^{7}=(-6 + 2\epsilon-7)^{7}=(-13 + 2\epsilon)^{7}<0\) (since \(-13+2\epsilon<0\) for small \(\epsilon\)), so \(f(x)=(2x - 7)^{7}(x + 3)^{4}<0\) (negative times positive is negative). So the graph touches the x - axis at \(x=-3\) and turns around.

Answer:

The x - intercepts of the function \(f(x)=(2x - 7)^{7}(x + 3)^{4}\) are \(x=\frac{7}{2}\) and \(x = - 3\).

• At \(x=\frac{7}{2}\), since the exponent of the factor \((2x - 7)\) is odd, the graph crosses the x - axis.
• At \(x=-3\), since the exponent of the factor \((x + 3)\) is even, the graph touches the x - axis and turns around.