Question

How do you change $y = - 2{x^2} + 8x - 1$ to vertex form?

Answer 1

Hint: In the equation above, we have an equation with two variables, and we are supposed to change it to vertex form. With the help of standard vertex form for a parabola formula, we can find the vertex values, which is
$y = a{(x - h)^2} + k$ with vertex $(h,k)$

Complete step-by-step solution:
Here we have an equation $y = - 2{x^2} + 8x - 1$ and we have to convert it to vertex form which can be done with the help of the formula,
We all know that the standard vertex form for a parabola is
$ \Rightarrow y = a{(x - h)^2} + k$ with vertex at $(h,k)$
When given a quadratic equation \[y = a{x^2} + bx + c\], the $x$ coordinate of the vertex is, $h = - \dfrac{b}{{2a}}$ and the $y$ coordinate of the vertex is $k = a{(h)^2} + b(h) + c$ then use the form $y = a{(x - h)^2} + k$
Explanation:
Applying the information in the answer to the given equation:
$ \Rightarrow h = - \dfrac{8}{{2( - 2)}}$
Multiplying with the bracket in denominator,
$ \Rightarrow h = - \dfrac{8}{{ - 4}}$
Therefore, on multiplying the negative signs and dividing the numerator with the denominator, we get,
$ \Rightarrow h = 2$
Substituting all values inside the formula, we get,
$ \Rightarrow k = - 2{(2)^2} + 8(2) - 1$
Substitute \[a = - 2,h = 2\], and \[k = 7\;\]into the form:
\[ \Rightarrow y = - 2{\left( {x - 2} \right)^2} + 7\]

Therefore, the equation $y = - 2{x^2} + 8x - 1$ will be converted into vertex form and be represented as \[y = - 2{\left( {x - 2} \right)^2} + 7\]

Note: The vertex form that we write is actually another form of writing out the equation of a parabola. The standard quadratic form is not really helpful while finding the vertex of a parabola. It needs to be converted to vertex form for that.
When written in "vertex form":
• $(h,k)$ is the vertex of the parabola, and $x = h$ is the axis of symmetry.
• The $h$ represents a horizontal shift (how far left, or right, the graph has shifted from $x = 0$).
• The $k$ represents a vertical shift (how far up, or down, the graph has shifted from $y = 0$).
• Notice that the $h$ value is subtracted in this form, and that the $k$ value is added.