Chapter 1.1, Problem 53E

Solution

To find: Axis of symmetry of the graph of a quadratic function that passes through two given points.

The axis of symmetry of the graph of a quadratic function is $x=-1$ .

Given data: The graph of quadratic function passes through points $\left(2,3\right),\left(-4,3\right)$ .

Method/Formula used:

A quadratic function $y\left(x\right)=a{\left(x-h\right)}^2+k$ represents a parabola with $x=h$ as the axis of symmetry, $\left(h,k\right)$ the vertex and ‘ a’ normal to the focus to that parabola. If $a<0$ , the parabola is inverted. For the vertex k is the height of parabola above x -axis.

Calculation:

Suppose the required quadratic function is $y\left(x\right)={\left(x-h\right)}^2+k$ that passes through points $\left(2,3\right),\left(-4,3\right)$ . Then its equation $y={\left(x-h\right)}^2+k$ must satisfy the points $\left(2,3\right),\left(-4,3\right)$ .

Substitute $x=2,y=3$ in the equation of the function, i.e.,

  $3={\left(2-h\right)}^2+k$ ... (1)

Substitute $x=-4,y=3$ in the equation of the function, i.e.,

  $3={\left(-4-h\right)}^2+k$

OR

  $\left.\begin{array}{l}3={\left(-1\right)}^2{\left[4+h\right]}^2+k\\ 3={\left(4+h\right)}^2+k\because {\left(-1\right)}^2=1\end{array}\right\}$ ... (2)

From equations (1) and (2),

  $\begin{array}{l}{\left(4+h\right)}^2+k={\left(2-h\right)}^2+k\\ {\left(4+h\right)}^2-{\left(2-h\right)}^2=k-k\\ {\left(4+h\right)}^2-{\left(2-h\right)}^2=0\end{array}$

Using formula, $a^2-b^2=\left(a-b\right)\left(a+b\right)$

  $\begin{array}{c}\left[\left(4+h\right)-\left(2-h\right)\right]\left[\left(4+h\right)+\left(2-h\right)\right]=0\\ \left[4+h-2+h\right]\left[4+h+2+h\right]=0\\ \left[2h+2\right]\left[6\right]=0\\ 2\cdot 6\left(h+1\right)=0\end{array}$

Since, $2\cdot 6=12\ne 0$ , therefore, $\left(h+1\right)=0\Rightarrow h=-1$ .

Therefore, the equation of the axis of symmetry is $x=-1$ .

In short, if $\left(p,k\right),\left(q,k\right)$ are the points on the graph of quadratic function, then the axis of symmetry of the parabola represented by the quadratic function is $x=\frac{p+q}{2}$ . Here, $p=2$ and $q=-4$ , therefore, axis of symmetry is $x=\frac{2-4}{2}=-1$ .