find all x - intercepts of the following function. write your answer or answers as coordinate points. be…

Explanation:

Step1: Set $y = f(x)=0$

Set $\frac{2x^{2}-13x - 24}{2x^{2}+3x}=0$. A rational - function is zero when its numerator is zero and its denominator is non - zero. First, solve the numerator equation $2x^{2}-13x - 24 = 0$.

Step2: Factor the quadratic equation

For the quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b=-13$, $c = - 24$), we use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. We can factor $2x^{2}-13x - 24$ as $(2x + 3)(x-8)=0$.

Step3: Solve for $x$ from the factored form

Set each factor equal to zero: If $2x+3 = 0$, then $2x=-3$, and $x=-\frac{3}{2}$. If $x - 8=0$, then $x = 8$.

Step4: Check the denominator

The denominator is $2x^{2}+3x=x(2x + 3)$. When $x=-\frac{3}{2}$, the denominator $2x^{2}+3x=2\times(-\frac{3}{2})^{2}+3\times(-\frac{3}{2})=2\times\frac{9}{4}-\frac{9}{2}=\frac{9}{2}-\frac{9}{2}=0$. So $x = -\frac{3}{2}$ is not an $x$-intercept. When $x = 8$, the denominator $2x^{2}+3x=2\times8^{2}+3\times8=2\times64 + 24=128+24 = 152\neq0$.

Answer:

$(8,0)$