Chapter 2.1, Problem 40E

To determine

To write:The standard form of quadratic function that has indicated vertex and whose graph passes through given point, and verify the result using graphing utility.

Answer to Problem 40E

The standard form of the described quadratic function is $f\left(x\right)=-1\cdot {\left(x+\frac{1}{4}\right)}^2-1$ .

Explanation of Solution

Given data: $y$

The vertex of the graph of quadratic function (a parabolic curve) is $\left(-\frac{1}{4},-1\right)$ and the graph passes through the point $\left(0,-\frac{17}{16}\right)$ .

Concept used:

Everyquadratic function represents a parabola. Thestandard 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 standard quadratic function that represents a parabola with vertex $\left(\text{α,β}\right)$ is given by

  $f\left(x\right)=a{\left(x-\text{α}\right)}^2+\text{β}$  .........(1)

The vertex of given parabola is $\left(\text{α,β}\right)=\left(-\frac{1}{4},-1\right)\Rightarrow \text{α}=-\frac{1}{4},\text{β=}-1$

Substituting values of $\text{α}$ and $\text{β}$ in equation (1) we have

  $f\left(x\right)=a{\left(x+\frac{1}{4}\right)}^2-1$  .........(2)

Given parabola passes through point $\left(0,-\frac{17}{16}\right)$ , therefore, from equation (2)

  $f\left(0\right)=a{\left(0+\frac{1}{4}\right)}^2-1$

  $\Rightarrow -\frac{17}{16}=\frac{1}{16}a-1$$f\left(0\right)=-\frac{17}{16}$

  $\Rightarrow -\frac{17}{16}+1=\frac{1}{16}a$

  $\Rightarrow a=-1$

Substituting $a=-1$ in equation (2) the standard form of the given parabola is $f\left(x\right)=-1\cdot {\left(x+\frac{1}{4}\right)}^2-1$ .

Thus, the quadratic function $f\left(x\right)=-1\cdot {\left(x+\frac{1}{4}\right)}^2-1$ represents a parabola with vertex $\left(-\frac{1}{4},-1\right)$ and passing through point $\left(0,-\frac{17}{16}\right)$ .

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

Verification of the given description of the quadratic function:

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

  

From the graph, we observe that the graph of function $f\left(x\right)=-1\cdot {\left(x+\frac{1}{4}\right)}^2-1$ is downwards concave parabola having vertex at point $\left(-\frac{1}{4},-1\right)$ and passes through the point $\left(0,-\frac{17}{16}\right)$ that verify the given description of the quadratic function.

Conclusion:

The standard form of the described quadratic function is $f\left(x\right)=-1\cdot {\left(x+\frac{1}{4}\right)}^2-1$ .