Question

How do you solve \[y={{x}^{2}}-14x+24\] graphically and algebraically?

Answer 1

Hint: In this problem, we have to solve and find the value of x and y, graphically and algebraically. We know that to solve graphically, we have to find the x-intercept and y-intercept, where at x-intercept the value of y is 0 and at y-intercept the value of x is 0. To solve algebraically, we can take the quadratic equation given, we can solve the quadratic equation using quadratic formula to find the value of x and substitute the x values, to get the value of y.

Complete step by step answer:
We know that the given equation to be solved graphically and algebraically is,
\[y={{x}^{2}}-14x+24\]…… (1)
We can solve this algebraically.
We know that we can solve a quadratic equation by quadratic formula.
The quadratic formula for the equation \[a{{x}^{2}}+bx+c=0\] is,
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By comparing the given quadratic equation to the general equation, we get
a = 1, b = -14, c = 24,
we can substitute the values in quadratic formula, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{14\pm \sqrt{{{\left( -14 \right)}^{2}}-4\left( 24 \right)}}{2} \\
 & \Rightarrow x=\dfrac{14\pm \sqrt{196-96}}{2} \\
 & \Rightarrow x=\dfrac{14\pm \sqrt{100}}{2} \\
 & \Rightarrow x=\dfrac{24}{2},\dfrac{4}{2} \\
 & \Rightarrow x=12,2 \\
\end{align}\]
Now, we can substitute the x value in equation (1), we get
When x = 2, the value of y form equation (1) is
\[\begin{align}
  & \Rightarrow y={{2}^{2}}-14\left( 2 \right)+24 \\
 & \Rightarrow y=4-28+24 \\
 & \Rightarrow y=0 \\
\end{align}\]
When x = 12, the value of y from equation (1) is
\[\begin{align}
  & \Rightarrow y=144-168+24 \\
 & \Rightarrow y=0 \\
\end{align}\]
Therefore, the value of \[\left( x,y \right)=\left( 12,0 \right),\left( 2,0 \right)\].
We can now solve it graphically.
We know that at x-intercept y is 0, substituting 0 in equation (1), we get
\[\begin{align}
  & \Rightarrow 0={{x}^{2}}-14x+24 \\
 & \Rightarrow x=12,2 \\
\end{align}\]
The x-intercept is \[\left( 12,0 \right),\left( 2,0 \right)\]
We have already solved the above step algebraically.
We know that at y-intercept, x is 0, substituting 0 in equation (1), we get
\[\begin{align}
  & \Rightarrow y=0-0+24 \\
 & \Rightarrow y=24 \\
\end{align}\]
The y-intercept is \[\left( 0,24 \right)\]
 We can write the equation (1) as \[y={{\left( x-7 \right)}^{2}}-25\]
From this, we can say that the turning point is \[\left( 7,-25 \right)\].
We can see that graphically it is a u-shaped parabola that comes down through \[\left( 0,24 \right)\] then through \[\left( 2,0 \right)\], through the minimum \[\left( 7,-25 \right)\] an back up through \[\left( 12,0 \right)\].
We can plot the above point in the graph.

Note: Students make mistakes in finding the x-intercept and y-intercept. We should always remember that at x-intercept the value of y is 0 and at the y-intercept the value of x is 0. We can solve the quadratic equation using the quadratic formula to find the value of x and substitute the x values, to get the value of y.