Chapter 2.1, Problem 37E

To determine

To write:The equation of parabola in standard form, whose graph is given, and verify the result using graphing utility.

Answer to Problem 37E

The standard equation of the graphed parabola is $f\left(x\right)=-1\cdot {\left(x+1\right)}^2+4$ .

Explanation of Solution

Given data: $y$

The graph of the parabola is shown in Figure-1 here.

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:

Let the quadratic function $f\left(x\right)=a{x}^2+bx+c$ is the equation of the parabola shown in Figure-1. Since, the points $\left(-3,0\right)$ , $\left(-1,4\right)$ , and $\left(1,0\right)$ lie on the graph so they (points) satisfy the equation $y=a{x}^2+bx+c$ i.e.,

As the point $\left(-3,0\right)$ lies on the parabola

  $0=a{\left(-3\right)}^2+b\left(-3\right)+c$

  $\Rightarrow 9a-3b+c=0$  .........(1)

As the point $\left(-1,4\right)$ lies on the parabola

  $4=a{\left(-1\right)}^2+b\left(-1\right)+c$

  $\Rightarrow a-b+c=4$  .........(2)

As the point $\left(1,0\right)$ lies on the parabola

  $0=a{\left(1\right)}^2+b\left(1\right)+c$

  $a+b+c=0$  .........(3)

Now, subtracting equation (2) from equation (3)

  $2b=-4\Rightarrow b=-2$

Now, subtracting equation (2) from equation (1)

  $8a-2b=-4$

  $\Rightarrow 8a=2b-4=2\left(-2\right)-4=-8$

  $\Rightarrow a=-1$

From equation (3)

  $-1-2+c=0$

(Substituting values of a&b)

  $\Rightarrow c=3$

Thus, the equation of the graphed parabola is $y=-1x^2-2x+3$ .

The quadratic function that represents given parabola is $f\left(x\right)=-x^2-2x+3$ that can be written as

  $f\left(x\right)=-\left(x^2+2x+1\right)+4$

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

Equation (4) is the required standard form of the parabola.

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

Thus, the quadratic function $f\left(x\right)=-x^2-2x+3$ represents a parabola with vertex $\left(-1,4\right)$ .

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

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

  

We observe that the graph of function $f\left(x\right)=-x^2-2x+3$ is a downwards concave parabola having vertex $\left(-1,4\right)$ , exactly same as the given graph that verify the results described above.

Conclusion:

The standard equation of the graphed parabola is $f\left(x\right)=-1\cdot {\left(x+1\right)}^2+4$ .