Question

ap precalculus section 0.2 - parabolas with intercepts mini - quiz

1. consider the graph at the right, which of the following could be an equation for the graph?

(a) (f(x)=(x - 3)(x + 1))
(b) (f(x)=(x + 3)(x - 1))
(c) (f(x)=(x - 3)(x + 1)(x - 1))
(d) (f(x)=(x + 3)(x + 1)(x - 1))
consider the function (h(t)=t^{2}-7t + 12). which of the following is zero of function (h)?
(a) (t=-3)
(b) (t=\frac{12}{7})
(c) (t = 4)
(d) (t=-\frac{12}{7})

Explanation:

Step1: Find the x - intercepts for the first question

The x - intercepts of a function \(y = f(x)\) are the values of \(x\) for which \(f(x)=0\). For a quadratic function in factored form \(y=a(x - r_1)(x - r_2)\), the x - intercepts are \(x = r_1\) and \(x = r_2\). From the graph, the x - intercepts are \(x=-1\) and \(x = 3\). For the function \(f(x)=(x - 3)(x + 1)\), when \(f(x)=0\), we set \((x - 3)(x + 1)=0\). Using the zero - product property \(x-3=0\) gives \(x = 3\) and \(x + 1=0\) gives \(x=-1\).

Step2: Find the zeros for the second question

To find the zeros of the function \(h(t)=t^{2}-7t + 12\), we set \(h(t)=0\), so \(t^{2}-7t + 12=0\). Factor the quadratic equation: \(t^{2}-7t + 12=(t - 3)(t - 4)=0\). Using the zero - product property, if \((t - 3)(t - 4)=0\), then \(t-3=0\) or \(t - 4=0\), which gives \(t = 3\) or \(t = 4\).

Answer:

1. A. \(f(x)=(x - 3)(x + 1)\)
2. C. \(t = 4\)