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Chapter 2: Problem 63

Use a graphing utility to graph the quadratic function. Find the \(x\) -intercept(s) of the graph and compare them with the solutions of the corresponding quadratic equation when \(f(x)=0\). $$f(x)=2 x^{2}-7 x-30$$

Short Answer

Expert verified
The solutions to the equation \(2x^2 - 7x - 30 = 0\), or the x-intercepts of the function \(f(x) = 2x^2 - 7x - 30\), are \(x = 5\) and \(x = -3\).

Step by step solution

01

Graphing the function

Start by graphing the function \(f(x) = 2x^2 - 7x - 30\). Typically, a quadratic function will give a parabola-shaped graph.
02

Find the x-intercepts

The x-intercepts are the values of \(x\) where the function equals zero (meaning, where it intersects the x-axis). Inspect these points on the graph.
03

Solve the quadratic equation

To verify if our graphical solution is correct, we'll solve the corresponding quadratic equation when \(f(x) = 0\). The equation for this will be \(2x^2 - 7x - 30 = 0\). This can be solved either factoring, completing the square or by using the quadratic formula \((-b 卤 \sqrt{b^2 - 4ac}) / 2a\).
04

Compare the solutions

Once you've found the solutions for \(x\) in Step 3, compare them with the x-intercepts identified from the graph in Step 2, they should be the same.

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Most popular questions from this chapter

Match the cubic function with the numbers of rational and irrational zeros. (a) Rational zeros: \(0 ;\) irrational zeros: 1 (b) Rational zeros: \(3 ;\) irrational zeros: 0 (c) Rational zeros: \(1 ;\) irrational zeros: 2 (d) Rational zeros: \(1 ;\) irrational zeros: 0 $$f(x)=x^{3}-x$$

Find the rational zeros of the polynomial function. $$f(z)=z^{3}+\frac{11}{6} z^{2}-\frac{1}{2} z-\frac{1}{3}=\frac{1}{6}\left(6 z^{3}+11 z^{2}-3 z-2\right)$$

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$f(x)=4 x^{3}-3 x^{2}+2 x-1$$

Find all the zeros of the function. When there is an extended list of possible rational zeros, use a graphing utility to graph the function in order to disregard any of the possible rational zeros that are obviously not zeros of the function. $$f(x)=16 x^{3}-20 x^{2}-4 x+15$$

Think About It \(\quad\) Sketch the graph of a fifth-degree polynomial function whose leading coefficient is positive and that has a zero at \(x=3\) of multiplicity 2
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