Chapter 2.1, Problem 34E

To determine

To describe:The graph of given quadraticfunction, identify the vertex, find the x- intercept and verify the results with graphing utility.

Explanation of Solution

Given data: $y$

The given quadratic function is $f\left(x\right)=-\left(x^2+3x-4\right)$ .

Concept used:

The standard form of quadratic function $f\left(x\right)=a{\left(x-\text{α}\right)}^2+\text{β}$ represents a parabola havingvertex $\left(\text{α,β}\right)$ . If $a>0$ then the parabola has minima at the vertex, so concave upwards, and if $a<0$ , then the parabola has maxima at the vertex so concave downwards

Calculations:

The given quadratic function is $f\left(x\right)=-\left(x^2+3x-4\right)$ may be written as

  $f\left(x\right)=-\left(x^2+3x-4\right)=-\left(x^2+2\cdot \frac{3}{2}\cdot x+\frac{9}{4}-4-\frac{9}{4}\right)$

  $f\left(x\right)=-\left(x^2+2\cdot \frac{3}{2}\cdot x+\frac{9}{4}\right)+\frac{25}{4}$

  $\Rightarrow f\left(x\right)=-1\cdot {\left(x+\frac{3}{2}\right)}^2+\frac{25}{4}$  .........(1)

Now, comparing equation (1) with the function $f(x)=a{\left(x-\text{α}\right)}^{\text{2}}\text{+β}$ , we have $a\text{=-}1$ , $\text{α=-}\frac{3}{2}$ , and $\text{β}=\frac{25}{4}$ .

Thus, the quadratic function $f\left(x\right)=-\left(x^2+3x-4\right)$ represents a parabola with vertex $\left(-\frac{3}{2},\frac{25}{4}\right)$ .

Since, $a=-1<0$ , hence the parabola has maximum at the vertex so is concave downwards.

The x-intercept of the parabola is given by substituting $f\left(x\right)=0$ and solving this equation for variable “x”.

  $f\left(x\right)=0\Rightarrow -1\cdot {\left(x+\frac{3}{2}\right)}^2+\frac{25}{4}=0$

  $\Rightarrow {\left(x+\frac{3}{2}\right)}^2=\frac{25}{4}$

  $\Rightarrow \left(x+\frac{3}{2}\right)=\pm \sqrt{\frac{25}{4}=}\pm \frac{5}{2}$

  $\Rightarrow x=-\frac{3}{2}+\frac{5}{2},-\frac{3}{2}-\frac{5}{2}$

  $\Rightarrow x=1,-4$ .

The curve parabola intersects x-axis at points $\left(1,0\right)$ , and $\left(-4,0\right)$ . Therefore, the intercepts of parabola on x-axis are $OA=1$ and $OB=-4$ .

The graph of the function $f\left(x\right)=-\left(x^2+3x-4\right)$ (a parabola) is shown in Figure-1 here.

  

We observe that the graph of function $f\left(x\right)=-\left(x^2+3x-4\right)$ is an upwards concave parabola having vertex $\left(-\frac{3}{2},\frac{25}{4}\right)$ and two intercepts $OA=1$ and $OB=-4$ that verify the results described above The graph of the function $f\left(x\right)=-\left(x^2+3x-4\right)$ is by(i) reflecting the graph of function $h\left(x\right)=x^2$ through x-axis, then (ii) shifting through a distance 25/4 units alongy-axis, and then (ii) shifting it through distance3/2 unitsparallel tox−axis left to the origin.