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Question: How do I convert the equation \(f\left( x \right) = {x^2} - 2x - 3\) to vertex form?...

How do I convert the equation f(x)=x2−2x−3f\left( x \right) = {x^2} - 2x - 3 to vertex form?

Explanation

Solution

We have to convert the equation f(x)=x2−2x−3f\left( x \right) = {x^2} - 2x - 3 to vertex form. For this, complete the square for x2−2x−3{x^2} - 2x - 3. Use the form ax2+bx+ca{x^2} + bx + c, to find the values of aa, bb, and cc. Consider the vertex form of a parabola to be (I). Next, substitute the values of aa and bb into the formula (II) and simplify the right side. Next, find the value of ee using the formula (III). Next, substitute the values of aa, dd, and ee into the vertex form (I). Finally, set yy equal to the new right side and get the required result.
Formula used:
I). Vertex form of a parabola: a(x+d)2+ea{\left( {x + d} \right)^2} + e
II). d=b2ad = \dfrac{b}{{2a}}
III). e=c−b24ae = c - \dfrac{{{b^2}}}{{4a}}
IV). Vertex form: y=a(x−h)2+ky = a{\left( {x - h} \right)^2} + k
V). Vertex: (h,k)\left( {h,k} \right)
VI). p=14ap = \dfrac{1}{{4a}}
VII). Focus: (h,k+p)\left( {h,k + p} \right)
VIII). Directrix: y=k−py = k - p

Complete step-by-step solution:
We have to convert the equation f(x)=x2−2x−3f\left( x \right) = {x^2} - 2x - 3 to vertex form.
For this, complete the square for x2−2x−3{x^2} - 2x - 3.
Use the form ax2+bx+ca{x^2} + bx + c, to find the values of aa, bb, and cc.
a=1,b=−2,c=−3a = 1,b = - 2,c = - 3
Consider the vertex form of a parabola.
a(x+d)2+ea{\left( {x + d} \right)^2} + e
Now, substitute the values of aa and bb into the formula d=b2ad = \dfrac{b}{{2a}}.
d=−22×1d = \dfrac{{ - 2}}{{2 \times 1}}
Simplify the right side.
⇒d=−1\Rightarrow d = - 1
Find the value of ee using the formula e=c−b24ae = c - \dfrac{{{b^2}}}{{4a}}.
e=−3−(−2)24×1e = - 3 - \dfrac{{{{\left( { - 2} \right)}^2}}}{{4 \times 1}}
⇒e=−4\Rightarrow e = - 4
Now, substitute the values of aa, dd, and ee into the vertex form a(x+d)2+ea{\left( {x + d} \right)^2} + e.
(x−1)2−4{\left( {x - 1} \right)^2} - 4
Set yy equal to the new right side.
y=(x−1)2−4y = {\left( {x - 1} \right)^2} - 4
Hence, the vertex form of the given function is y=(x−1)2−4y = {\left( {x - 1} \right)^2} - 4.

Note: We can also convert the equation f(x)=x2−2x−3f\left( x \right) = {x^2} - 2x - 3 to vertex form by directly completing the square.
Vertex form can be represented as y=(x−h)2+ky = {\left( {x - h} \right)^2} + k
where the point (h,k)\left( {h,k} \right) is the vertex.
To do that, we should complete the square of
y=x2−2x−3y = {x^2} - 2x - 3
First, we should try to change the last number in a way so we can factor the entire thing
⇒ we should aim for y=x2−2x+1y = {x^2} - 2x + 1
to make it look like y=(x−1)2y = {\left( {x - 1} \right)^2}
If we notice, the only difference between the original y=x2−2x−3y = {x^2} - 2x - 3 and the factor-able y=x2−2x+1y = {x^2} - 2x + 1 is simply changing the −3 - 3 to 11.
Since we can't randomly change the −3 - 3 to a 1, we can add 1 and subtract a 1 to the equation at the same time to keep it balanced.
So, we get
y=x2−2x+1−3−1y = {x^2} - 2x + 1 - 3 - 1
Organizing
y=(x2−2x+1)−3−1y = \left( {{x^2} - 2x + 1} \right) - 3 - 1
Add like terms
−3−1=−4- 3 - 1 = - 4
Factor
y=(x−1)2−4y = {\left( {x - 1} \right)^2} - 4
Hence, the vertex form of the given function is y=(x−1)2−4y = {\left( {x - 1} \right)^2} - 4.