Question

201 - ma1 - dw calculus i - social science, math department, dawson college

1. consider

$f(x) = \dfrac{x^2 - 3x - 4}{x^2}$
$f(x) = \dfrac{3x + 8}{x^3}$
$f(x) = \dfrac{-6x - 24}{x^4}$
a. 1 mark find the x-intercept(s) of f(x)
$f(x) = \dfrac{(x + 1)(x - 4)}{}$

Explanation:

Step1: Set \( f(x) = 0 \)

To find the \( x \)-intercepts, we set \( f(x)=\frac{x^{2}-3x - 4}{x^{2}}=0 \). A fraction is zero when its numerator is zero (and the denominator is not zero). So we solve \( x^{2}-3x - 4 = 0 \) and check \( x^{2}
eq0 \).

Step2: Factor the numerator

Factor \( x^{2}-3x - 4 \). We need two numbers that multiply to - 4 and add to - 3. The numbers are - 4 and 1. So \( x^{2}-3x - 4=(x - 4)(x+1) \).

Step3: Solve for \( x \)

Set \( (x - 4)(x + 1)=0 \). Using the zero - product property, we have \( x-4 = 0 \) or \( x + 1=0 \). So \( x = 4 \) or \( x=-1 \).

Step4: Check the denominator

For \( x = 4 \), the denominator \( x^{2}=16
eq0 \). For \( x=-1 \), the denominator \( x^{2}=1
eq0 \).

Answer:

The \( x \)-intercepts are \( x=-1 \) and \( x = 4 \) (or the points \((-1,0)\) and \((4,0)\))