Chapter 1, Problem 15TP

Solution

To find: The $x$ coordinate of the vertex of the parabola that passes through the point $\left(0,-22\right),\left(2,-6\right),\left(5,12\right)$ .

The $x$ coordinate of the vertex of the parabola is $3.$

Given information:

The parabola passing through the points are $\left(0,-22\right),\left(2,-6\right),\left(5,12\right)$ .

Concept used:

The equation of the parabola is of the form $y=a{x}^2+bx+c$ .

Vertex of the $x$ coordinate of the parabola $x=-\frac{b}{2a}$ .

Calculations:

If the parabola is passing through the point $\left(0.-22\right)$ ,then equation of parabola becomes:

  $\begin{array}{l}-22=c\\ c=-22\end{array}$

Now the parabola passing through the point $\left(2,-6\right)$ then the equation becomes:

  $-6=4a+2b+c$

Substitute the the value of $c=-22$

  $\begin{array}{l}4a+2b=22-6\\ 4a+2b=16\\ 2a+b=8\left(1\right)\end{array}$

Now the parabola passing through the point $\left(5,-12\right)$ then the equation becomes:

  $-12=25a+5b+c$

Substitute the value of $c=-22$

  $\begin{array}{l}25a+5b=22-12\\ 25a+5b=10\\ 5a+b=2\left(2\right)\end{array}$

Now solving the equation $\left(1\right)\text{and}\left(2\right).$

  $\begin{array}{l}2a+b=8\left(1\right)\\ 5a+b=2\left(2\right)\\ \left(1\right)-\left(2\right)\Rightarrow -3a=6\\ \Rightarrow a=-2\\ \text{Substitute}a=-2\text{in}\left(1\right)\\ b=12\end{array}$

Substitute values of $a=-2,b=12,c=-22$ in the equation of the parabola:

  $y=-2x^2+12x-22$ .

Hence $x$ coordinate of the vertex of the parabola:

  $\begin{array}{c}x=\frac{-12}{-2\times 2}\\ =3\end{array}$

The $x$ coordinate of the vertex of the parabola is $3.$ c