Question

question 15 of 22. if the function f(x)=2x² - 7x + 5 is in the xy - plane, where y = f(x), then what is the value of the x - intercept of the function f(x)? a. (-⅖,0) b. (-1,0) c. (0,⅖) d. (⅖,0)

Explanation:

Step1: Recall x - intercept definition

The x - intercept is found by setting \(y = f(x)=0\). So we set \(2x^{2}-7x + 5=0\).

Step2: Factor the quadratic equation

We factor \(2x^{2}-7x + 5\) as \(2x^{2}-2x-5x + 5=2x(x - 1)-5(x - 1)=(2x - 5)(x - 1)=0\).

Step3: Solve for x

If \(2x-5 = 0\), then \(2x=5\), \(x=\frac{5}{2}\); if \(x - 1=0\), then \(x = 1\). The x - intercepts are of the form \((x,0)\), so the x - intercepts are \((1,0)\) and \((\frac{5}{2},0)\). But if we assume there is a mistake in the problem - setup and we use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c=0\) (here \(a = 2\), \(b=-7\), \(c = 5\)). \(x=\frac{7\pm\sqrt{(-7)^{2}-4\times2\times5}}{2\times2}=\frac{7\pm\sqrt{49 - 40}}{4}=\frac{7\pm\sqrt{9}}{4}=\frac{7\pm3}{4}\). When we take the plus - sign \(x=\frac{7 + 3}{4}=\frac{10}{4}=\frac{5}{2}\), when we take the minus - sign \(x=\frac{7-3}{4}=1\).

Answer:

None of the given options are correct. If we assume there is a mis - typing and we consider the general quadratic solution process, the x - intercepts of \(y = 2x^{2}-7x + 5\) are \((1,0)\) and \((\frac{5}{2},0)\).