Question

How do you graph $ {(y - 3)^2} = 8(x - 3) $

Answer 1

Hint: In this question we are given an equation in terms of x and y and we have to find its graph. To find the graph of an equation, we find the coordinates of some points lying on the curve of the given equation. So, to find the coordinates of the points lying on the graph of the given equation, we will express one of the variables in terms of the other variable. Let, y be the dependent variable so we will rearrange the equation and express y in terms of x.

Complete step-by-step answer:
We have to graph $ {(y - 3)^2} = 8(x - 3) $
We will rearrange this equation as –
 $
  {(y - 3)^2} = 8x - 24 \\
   \Rightarrow y - 3 = \pm \sqrt {8x - 24} \\
   \Rightarrow y = 3 \pm \sqrt {8x - 24} \;
  $
For y to be defined
 $
  8x - 24 > 0 \\
   \Rightarrow 8x > 24 \\
   \Rightarrow x > 3 \;
  $
So, the plot of this function will contain values of x greater than 3.
At $ x = 3,\,y = 3 \pm \sqrt {8(3) - 24} = 3 $
The given parabolic function has vertex at $ (3,3) $ .
 $
  y = 0,\,0 = 3 \pm \sqrt {8x - 24} \\
   \Rightarrow \sqrt {8x - 24} = 3,\,\sqrt {8x - 24} = - 3 \;
  $
We know that $ x - 3 > 0 $ so $ \sqrt {x - 3} $ cannot be negative. So, we get –
 $
  8x - 24 = 9 \\
   \Rightarrow 8x = 35 \\
   \Rightarrow x = \dfrac{{35}}{8} = 4.375 \;
  $
So, we have got that the given parabola has a vertex at $ (3,3) $ , has an x-intercept at $ (4.375,0) $ and x can take values greater than 3 only. Thus, the graph of this function is given as –

Note: In the given equation, the degree of one of the variables is 2, so the graph of this equation will be parabolic, that is, the graph of this equation will be a U-shaped curve. As the degree of y is 2, so the vertex of this parabola lies on the x-axis. To find the coordinates, we put random values of one variable and get the values of the other variable from the given equation.