Chapter 2.1, Problem 32E

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 $f\left(x\right)=-4x^2+24x-41$ .

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 $f\left(x\right)=-4x^2+24x-41$ may be written as

  $f\left(x\right)=-4x^2+24x-41$

  $f\left(x\right)=-4x^2+24x-36-5$

  $f\left(x\right)=-4\left(x^2-6x+9\right)-5$

  $\Rightarrow f\left(x\right)=-4\cdot {\left(x-3\right)}^2-5$  .........(1)

Now, comparing equation (1) with the function $f(x)=a{\left(x-\text{α}\right)}^{\text{2}}\text{+β}$ , we have $a\text{=-}4$ , $\text{α=}3$ , and $\text{β}=-5$ .

Thus, the quadratic function $f\left(x\right)=-4x^2+24x-41$ represents a parabola with vertex $\left(3,-5\right)$ .

Since, $a=-4<0$ , 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”.

  $\Rightarrow f\left(x\right)=0\Rightarrow -4\cdot {\left(x-3\right)}^2-5=0$

  $\Rightarrow {\left(x-3\right)}^2=-\frac{5}{4}$

  $\Rightarrow \left(x-3\right)=\pm \frac{\sqrt{-5}}{2}$ , that is an imaginary value.

Therefore, the curve of function $f\left(x\right)=-4x^2+24x-41$ does not intersect x-axis.

Hence, there is no intercept made by the graph of function $f\left(x\right)=-4x^2+24x-41$ .

The graph of the function $f\left(x\right)=-4x^2+24x-41$ (a parabola) is shown in Figure-1 here.

  

We observe that the graph of function $f\left(x\right)=-4x^2+24x-41$ is a downwards concave parabola having vertex $\left(3,-5\right)$ and there is no intercept of the parabola on x- axisthat verify the results described above.

The graph of the function $f\left(x\right)=-4x^2+24x-41$ is by(i) reflecting the graph of function $h\left(x\right)=x^2$ through x-axis, then (ii) shifting through a distance5 units along negativey-axis, and then (ii) shifting it through distance3unitsparallel tox−axisright to the origin.