Chapter 2, Problem 6CT

To determine

To calculate: Equation of the Parabola

Answer to Problem 6CT

The equation of parabola is: $y=-\frac{1}{4}{(x-3)}^2+4$

Explanation of Solution

Given information: Three points in the graph are $(-1,0),(5,3),(7,0)$

Calculation:

Consider the general quadratic equation:

  $y=a{x}^2+bx+c$

Substitute the given points in the general equation

  $\begin{array}{l}(x,y)=(-1,0)\\ $ 0=a{(-1)}^2+b(-1)+c$=>0=a-b+c\mathrm{...}(i)\\ (x,y)=(5,3)\\ $ 3=a{(5)}^2+b(5)+c$=>3=25a+5b+c\mathrm{...}(ii)\\ (x,y)=(7,0)\\ $ 0=a{(7)}^2+b(7)+c$=>0=49a+7b+c\mathrm{...}(iii)\end{array}$

Solve the three equations in three variables:

Subtract $(i)$ from $(ii)$ and $(i)$ from $(iii)$

  $\begin{array}{l}3=24a+6b\mathrm{...}(iv)\\ 0=48a+8b\mathrm{...}(v)\end{array}$

Divide $(iv)$ by 3 and $(v)$ by 8

  $\begin{array}{l}1=8a+2b\mathrm{...}(vi)\\ 0=6a+b\mathrm{...}(vii)\\ =>b=-6a\end{array}$

Substitute this in $(vi)$

  $\begin{array}{l}1=8a+2(-6a)\\ =>1=8a-12a\\ =>1=-4a\\ =>a=-\frac{1}{4}\\ b=-6a=>b=-6\left(-\frac{1}{4}\right)=\frac{3}{2}\end{array}$

Substitute these values in $(i)$

  $\begin{array}{l}0=-\frac{1}{4}-\frac{3}{2}+c\\ =>c=\frac{1}{4}+\frac{3}{2}=\frac{1+6}{4}=\frac{7}{4}\end{array}$

Thus, the equation of parabola is given by:

  $\begin{array}{l}y=-\frac{1}{4}{x}^2+\frac{3}{2}x+\frac{7}{4}\\ =>4y=-x^2+6x+7\\ =>4y=-x^2+6x-9+7+9\\ =>4y=-(x^2-6x+9)+16\\ $ =>4y=-{(x-3)}^2+16$$ =>y=-\frac{1}{4}{(x-3)}^2+4$\end{array}$