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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.

**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.

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