Question

How to graph a parabola $y = {\left( {x + 5} \right)^2} - 3$?

Answer 1

Hint: First, find the x-intercepts of the curve by putting $y = 0$. After that, find the y-intercept by putting $x = 0$. Then take a minimum of 5 points and plot the points. After plotting the points, join the points with a smooth freehand curve and identify the curve that we have obtained.

Complete step by step solution:
We know that the graph of a function is the locus of points $\left( {x,y} \right)$ such that $y = f\left( x \right)$ where x, y are real numbers. We are given the following quadratic polynomial function,
$ \Rightarrow y = {\left( {x + 5} \right)^2} - 3$
So, let us put $y = 0$ and find the x-intercept. We have,
$ \Rightarrow 0 = {\left( {x + 5} \right)^2} - 3$
Move 3 on the other side,
$ \Rightarrow {\left( {x + 5} \right)^2} = 3$
Take the square root on both sides,
$ \Rightarrow x + 5 = \pm \sqrt 3 $
Subtract 5 on both sides,
$ \Rightarrow x + 5 - 5 = \pm \sqrt 3 - 5$
Simplify the terms,
$ \Rightarrow x = - 5 \pm \sqrt 3 $
It means the curve cuts the x-axis at $\left( { - 5 + \sqrt 3 ,0} \right)$ and $\left( { - 5 - \sqrt 3 ,0} \right)$.
Let us put $x = 0$ and find the y-intercept. We have,
$ \Rightarrow y = {\left( {0 + 5} \right)^2} - 3$
Simplify the terms,
$ \Rightarrow y = {\left( 5 \right)^2} - 3$
Square the term on the right side,
$ \Rightarrow y = 25 - 3$
Simplify the terms,
$ \Rightarrow y = 22$
It means the curve cuts the y-axis at $\left( {0,22} \right)$.
We know that all quadratic functions of the type $y = a{x^2} + bc + c$ have minimum values but not maximum.
Since the square is always non-negative, we have ${\left( {x + 5} \right)^2} \ge 0$, then we have
$ \Rightarrow y = {\left( {x + 5} \right)^2} - 3 \ge 0$
So, the minimum value of $y = - 3$ and the minimum value occurs when ${\left( {x + 5} \right)^2} = 0$ or $x = - 5$.
We have already three points for the curve $\left( { - 5 - \sqrt 3 ,0} \right)$, $\left( { - 5 + \sqrt 3 ,0} \right)$ and $\left( {0,22} \right)$. We find y for two more points.
At $x = - 5$ we have,
$ \Rightarrow y = {\left( { - 5 + 5} \right)^2} - 3$
Simplify the terms,
$ \Rightarrow y = {\left( 0 \right)^2} - 3$
Simplify the terms,
$ \Rightarrow y = - 3$
At $x = - 2$ we have,
$ \Rightarrow y = {\left( { - 2 + 5} \right)^2} - 3$
Simplify the terms,
$ \Rightarrow y = {\left( 3 \right)^2} - 3$
Square the term on the right side,
$ \Rightarrow y = 9 - 3$
Simplify the terms,
$ \Rightarrow y = 6$
So, we draw the table for x and y.

$X$	$ - 5 - \sqrt 3 $	$ - 5 + \sqrt 3 $	$0$	$-5$	$-2$
$Y$	$0$	$0$	$22$	$-3$	$6$

We plot the above points and join them to have the graph as

Note: We note that the obtained graph is the graph of the upward parabola whose general equation is given by $y = a{x^2} + bx + c$ with the condition $a > 0$ whose vertex here is $\left( { - 5, - 3} \right)$ . We can directly find the minimum value of $y = {\left( {x + 5} \right)^2} - 3$ by finding $x = - \dfrac{b}{{2a}}$. If $a < 0$ the equation $y = a{x^2} + bx + c$ represents a downward parabola. We also note that the obtained curve is symmetric about the line $x = - 5$.