Chapter 2.1, Problem 53E

To determine

To find: two quadratic functions, one that open upward and one that opens downward.

Answer to Problem 53E

One that opens upward, $2x^2+7x+3$

One that open downward, $-2x^2-7x-3$

Explanation of Solution

Given information: x-intercepts of quadratic function is given as follows-

  $(-3,0),(\frac{-1}{2},0)$.

The x-intercepts of quadratic function are $(-3,0),(\frac{-1}{2},0)$.

Then zeroes of quadratic function are $\text{-3 and }\frac{-1}{2}$

Sum of zeroes = $-3+(-\frac{1}{2})=\frac{-6-1}{2}=-\frac{7}{2}$

Product of zeroes = $-3\times (-\frac{1}{2})=\frac{3}{2}$

Then quadratic equation whose zeroes are $\text{-3 and }\frac{-1}{2}$ is given by

  $x^2-x(\text{sum of zeroes)+(product of zeroes)=0}$

  $\begin{array}{l}\Rightarrow {\text{x}}^2-x(-\frac{7}{2})+\frac{3}{2}=0\\ \Rightarrow x^2+\frac{7}{2}x+\frac{3}{2}=0\end{array}$

  $\begin{array}{l}\Rightarrow \frac{2x^2+7x+3}{2}=0\\ \Rightarrow 2x^2+7x+3=0\end{array}$

Hence required quadratic function that open upward is $2x^2+7x+3$ .

Here coefficient of $x^2$ is positive so it is open upward,

To find a quadratic function which is open downward multiply the equation by ‘−‘

  $\Rightarrow -2x^2-7x-3=0$

Hence required quadratic function that open downward is $-2x^2-7x-3$ .