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# How do you solve $y={{x}^{2}}-14x+24$ graphically and algebraically?

Last updated date: 24th Jul 2024
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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.

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.