Chapter 2.2, Problem 11E

To determine

To identify (match):The graph of a polynomial from the given list of graphs.

Answer to Problem 11E

The polynomial function $f\left(x\right)=-2x^2-5x$ represents the graph (c) from the list.

Explanation of Solution

Given information:

The given polynomial function is $f\left(x\right)=-2x^2-5x$ .

Concept used:

The graph of standard form of quadratic function $f\left(x\right)=a{\left(x-h\right)}^2+k$ is a parabola with vertex $\left(h,k\right)$ . If $a>0$ the parabola is concave-upwards.

The x- intercept of the graph of quadratic function $f\left(x\right)$ is $\left(a,0\right)$ , where $f\left(a\right)=0$ .

The given quadratic polynomial $f\left(x\right)=-2x^2-5x$ can be written as:

  $f\left(x\right)=-2\left(x^2+\frac{5}{2}x\right)$

  $f\left(x\right)=-2\left(x^2+2\cdot \frac{5}{4}x+\frac{25}{16}\right)+\frac{25}{8}$

  $\Rightarrow f\left(x\right)=-2{\left(x+\frac{5}{4}\right)}^2+\frac{25}{8}$

  $\Rightarrow f\left(x\right)=-2{\left(x+1.25\right)}^2+3.125$...............(1)

The vertex of the parabola represented by $f\left(x\right)=-2x^2-5x$ is $\left(-1.25,3.125\right)$ . Since, $a=-2<0$ so the parabola opening downwards.

Now, $f\left(x\right)=0\Rightarrow -2x\left(x+\frac{5}{2}\right)=0$

  $\Rightarrow x\left(x+\frac{5}{2}\right)=0\Rightarrow x=0,x=-2.5$

Hence, the x-intercepts of the graph of $f\left(x\right)=-2x^2-5x$ are $\left(0,0\right)$ and $\left(-2.5,0\right)$ .

We see that the graph (c) is a parabola with vertex $\left(-1.25,3.125\right)$ that opens downwards and has $\left(0,0\right)$ and $\left(-2.5,0\right)$ as its x-intercepts.

  

Hence, the quadratic polynomial function $f\left(x\right)=-2x^2-5x$ represents the graph (c) from the list.

Conclusion:

The quadratic polynomial function $f\left(x\right)=-2x^2-5x$ represents the graph (c) from the list.