Question

Working on how to plot the axis of symmetry and the vertex for the function:h(x)=(x-5)^2-7

Answer 1

A generic expression of a quadratic is

`f(x)=ax^2+bx+c`

We can write it using the vertex form, that is

`f(x)=a(x-h)^2+k`

The vertex form holds a lot of important properties because it shows us immediately where the vertex is, just by looking at the value of "h" and "k" of the formula, in fact, the vertex of the parabola is

`(h,k)`

And the axis of symmetry of a parabola is the x-coordinate of the vertex, then, the axis of symmetry is

`x=h`

But how to identify h and k when we have the parabola in the vertex form? We have the following equation

`h(x)=(x-5)^2-7`

What's the value of the number that sums or subctract the quadratic term? In that case, it's -7, then it's the value of k

`k=-7`

Now to identify the "h" we must take care, it seems like h = -5 because the quadratic term is (x-5)² but we always change the signal of the number inside the quadratic term, if we have -5 inside it, the value of h is 5

`h=5`

Then, the vertex will be

`(h,k)=(5,-7)`

The vertex is (5, -7) and the axis of symmetry will be the same value of h, then

`\begin{gathered} x=h \\ \\ x=5 \end{gathered}`

Symmetry and vertex

`\begin{gathered} \text{ vertex: \lparen5, -7\rparen} \\ \\ \text{ axis of symmetry: x = 5} \end{gathered}`

Now, to plot the graph precisely we must find the roots of the parabola, in other words, the value of x that makes h(x) equal to zero:

`\begin{gathered} h(x)=0 \\ \\ (x-5)^2-7=0 \end{gathered}`

Then, we want to solve:

`(x-5)^2-7=0`

Put the quadratic term on one side

`\begin{gathered} (x-5)^2=7 \\ \end{gathered}`

Take the square root on both sides

`\begin{gathered} √((x-5)^2)=√(7) \\ \\ |x-5|=√(7) \end{gathered}`

Be careful! when we do the square root of the quadratic term we must remember to put the modulus. Then we will solve this modular equation:

`|x-5|=√(7)`

Which is the same as solving to different equations:

`|x-5|=√(7)\Rightarrow\begin{cases}x-5={√(7)} \\ x-5=-{√(7)}\end{cases}`

Then the two solutions are

`\begin{gathered} x=5+√(7)\approx7.65 \\ \\ x=5-√(7)\approx2.35 \end{gathered}`

Then we can do the plot of the parabola with a good precision

Or using a graphing calculator