Question

question 9:

• let ( f(x) = \frac{2x^2 - 11x + 15}{2x^2 - 5x + 3} )

find:
a.) x-intercept(s)
b.) y-intercept
c.) vertical asymptotes
d.) horizontal/slant asymptotes
e.) graph it!
graph grid omitted

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(2x^2 - 11x + 15\) and the denominator \(2x^2 - 5x + 3\).

For the numerator: \(2x^2 - 11x + 15\). We need two numbers that multiply to \(2\times15 = 30\) and add to \(-11\). Those numbers are \(-6\) and \(-5\). So,
\[ \begin{align*} 2x^2 - 11x + 15&=2x^2 - 6x - 5x + 15\\ &=2x(x - 3) - 5(x - 3)\\ &=(2x - 5)(x - 3) \end{align*} \]

For the denominator: \(2x^2 - 5x + 3\). We need two numbers that multiply to \(2\times3 = 6\) and add to \(-5\). Those numbers are \(-2\) and \(-3\). So,
\[ \begin{align*} 2x^2 - 5x + 3&=2x^2 - 2x - 3x + 3\\ &=2x(x - 1) - 3(x - 1)\\ &=(2x - 3)(x - 1) \end{align*} \]

So, \(f(x)=\frac{(2x - 5)(x - 3)}{(2x - 3)(x - 1)}\) (note that we should check for common factors, but there are none here).

Step2: Find x-intercepts (a)

To find the x-intercepts, set \(f(x) = 0\). A rational function is zero when its numerator is zero (and the denominator is not zero at those points).

Set the numerator equal to zero: \((2x - 5)(x - 3)=0\)

Solve for \(x\):
- \(2x - 5 = 0\) gives \(x=\frac{5}{2}\)
- \(x - 3 = 0\) gives \(x = 3\)

Now, check if these values make the denominator zero:
- For \(x=\frac{5}{2}\), denominator is \((2\times\frac{5}{2}-3)(\frac{5}{2}-1)=(5 - 3)(\frac{3}{2})=(2)(\frac{3}{2}) = 3 eq0\)
- For \(x = 3\), denominator is \((2\times3 - 3)(3 - 1)=(6 - 3)(2)=(3)(2)=6 eq0\)

So the x-intercepts are at \(x=\frac{5}{2}\) and \(x = 3\), so the points are \((\frac{5}{2},0)\) and \((3,0)\).

Step3: Find y-intercept (b)

To find the y-intercept, set \(x = 0\) and solve for \(y = f(0)\).

\(f(0)=\frac{2(0)^2 - 11(0)+15}{2(0)^2 - 5(0)+3}=\frac{15}{3}=5\)

So the y-intercept is at \((0,5)\).

Step4: Find vertical asymptotes (c)

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points).

Set the denominator equal to zero: \((2x - 3)(x - 1)=0\)

Solve for \(x\):
- \(2x - 3 = 0\) gives \(x=\frac{3}{2}\)
- \(x - 1 = 0\) gives \(x = 1\)

Check if the numerator is zero at these points:
- For \(x=\frac{3}{2}\), numerator is \((2\times\frac{3}{2}-5)(\frac{3}{2}-3)=(3 - 5)(-\frac{3}{2})=(-2)(-\frac{3}{2}) = 3 eq0\)
- For \(x = 1\), numerator is \((2\times1 - 5)(1 - 3)=(-3)(-2)=6 eq0\)

So the vertical asymptotes are at \(x=\frac{3}{2}\) and \(x = 1\).

Step5: Find horizontal/slant asymptotes (d)

To find horizontal or slant asymptotes, we compare the degrees of the numerator and denominator.

The degree of the numerator (highest power of \(x\)) is 2, and the degree of the denominator is also 2. When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.

The leading coefficient of the numerator is 2 (from \(2x^2\)) and the leading coefficient of the denominator is 2 (from \(2x^2\)). So the horizontal asymptote is \(y=\frac{2}{2}=1\).

Since the degrees are equal, there is no slant asymptote (slant asymptotes occur when the degree of the numerator is one more than the degree of the denominator).

Step6: Graphing (e)

To graph the function:
1. Plot the x-intercepts \((\frac{5}{2},0)\) and \((3,0)\) and the y-intercept \((0,5)\).
2. Draw the vertical asymptotes \(x = 1\) and \(x=\frac{3}{2}\) (dashed lines).
3. Draw the horizontal asymptote \(y = 1\) (dashed line).
4. Analyze the behavior of the function in each interval determined by the vertical asymptotes (\(x < 1\), \(1 < x < \frac{3}{2}\), \(\frac{3}{2}<x<\frac{5}{2}\), \(\frac{5}{2}<x<3\), \(x > 3\)) by picking test points and determining the sign of \(f(x)\) in each interval.

For example:
- For \(x < 1\), let's pick \(x = 0\), \(f(0)=5>0\), so the function is above the x-axis here.
- For \(1 < x < \frac{3}{2}\), let's pick \(x = 1.2\) (which is between 1 and 1.5). \(f(1.2)=\frac{(2\times1.2 - 5)(1.2 - 3)}{(2\times1.2 - 3)(1.2 - 1)}=\frac{(2.4 - 5)(-1.8)}{(2.4 - 3)(0.2)}=\frac{(-2.6)(-1.8)}{(-0.6)(0.2)}=\frac{4.68}{-0.12}=-39<0\), so the function is below the x-axis here.
- For \(\frac{3}{2}<x<\frac{5}{2}\), let's pick \(x = 2\). \(f(2)=\frac{(4 - 5)(2 - 3)}{(4 - 3)(2 - 1)}=\frac{(-1)(-1)}{(1)(1)} = 1>0\), so the function is above the x-axis here.
- For \(\frac{5}{2}<x<3\), let's pick \(x = 2.6\) (between 2.5 and 3). \(f(2.6)=\frac{(5.2 - 5)(2.6 - 3)}{(5.2 - 3)(2.6 - 1)}=\frac{(0.2)(-0.4)}{(2.2)(1.6)}=\frac{-0.08}{3.52}\approx - 0.0227<0\), so the function is below the x-axis here.
- For \(x > 3\), let's pick \(x = 4\). \(f(4)=\frac{(8 - 5)(4 - 3)}{(8 - 3)(4 - 1)}=\frac{(3)(1)}{(5)(3)}=\frac{3}{15}=0.2>0\), so the function is above the x-axis here.

Using this information, we can sketch the graph, approaching the vertical asymptotes and the horizontal asymptote.

Answers:

a.) The x-intercepts are \(\boldsymbol{(\frac{5}{2},0)}\) and \(\boldsymbol{(3,0)}\)

b.) The y-intercept is \(\boldsymbol{(0,5)}\)

c.) The vertical asymptotes are \(\boldsymbol{x = 1}\) and \(\boldsymbol{x=\frac{3}{2}}\)

d.) The horizontal asymptote is \(\boldsymbol{y = 1}\) (no slant asymptote)

e.) (Graph as described in Step6, with x-intercepts, y-intercept, vertical asymptotes at \(x = 1\), \(x=\frac{3}{2}\), horizontal asymptote at \(y = 1\), and the function's behavior in each interval as analyzed)

Answer:

s:
a.) The x-intercepts are \(\boldsymbol{(\frac{5}{2},0)}\) and \(\boldsymbol{(3,0)}\)

b.) The y-intercept is \(\boldsymbol{(0,5)}\)

c.) The vertical asymptotes are \(\boldsymbol{x = 1}\) and \(\boldsymbol{x=\frac{3}{2}}\)

d.) The horizontal asymptote is \(\boldsymbol{y = 1}\) (no slant asymptote)

e.) (Graph as described in Step6, with x-intercepts, y-intercept, vertical asymptotes at \(x = 1\), \(x=\frac{3}{2}\), horizontal asymptote at \(y = 1\), and the function's behavior in each interval as analyzed)