Difficulty turning a quadratic equation to "vertex"-form

Question

I'm having difficulty reducing a quadratic equation to its "vertex-form" by following my textbook and nearly every tutorial I can find online.

The starting equation is:

$$f(x) = -2x^2 + 16x - 24$$

Next, I divided each term by $-2$ (to leave the first term as $x^2$):

$$ = x^2 - 8x + 12$$

Next, to complete the square, I divide $x$'s coefficient by $2$, then square it ($(b/2)^2$):

$$\left(\frac{8}{2}\right)^2 = 16$$

Which gives me $16$. I add $16$ to complete the square, but also subtract it so it doesn't affect the value:

$$ = x^2 - 8x + 16 + 12 - 16$$

Finally, I complete the square:

$$ = (x - 4)^2 + 12 - 16$$

Then simplify:

$$= (x - 4)^2 - 4$$

According to by book though,the answer is:

$$-2(x - 4)^2 + 8$$

Unfortunately, the text skips over steps and makes it very unclear how they got the answer that they did. Can anyone point out where I went wrong above? I compiled the above steps from several videosand tutorials, but I must have gone wrong somewhere.

Answer 1

You made your error when you divided by $-2$. When you divided $f(x) = -2x^2 + 16x - 24$ by $-2$, you obtained $\color{red}{f(x) = x^2 - 8x + 12}$. The expression in red is not equal to the original function since the $y$-intercept of the original function is $-24$, while the $y$-intercept of the function in red is $12$.

You want to transform the equation $f(x) = -2x^2 + 16x - 24$ into the form $f(x) = a(x - h)^2 + k$. We first extract a factor of $-2$ from the quadratic and linear terms so that the expression in parentheses has the form $x^2 + 2kx$. We do this since we want to complete the square on $x^2 + kx$ to form the perfect square $x^2 + 2kx + k^2 = (x + k)^2$. \begin{align*} f(x) & = -2x^2 + 16x - 24\\ & = -2(x^2 - 8x) - 24 \end{align*} The term in parentheses now has the form $x^2 + 2kx$, where $2k = -8$, so $k = -4$, and $k^2 = 16$. If we add $16$ inside the parentheses, we will obtain the perfect square $x^2 - 8x + 16 = (x - 4)^2$. However, adding $16$ inside the parentheses adds $-2 \cdot 16 = -32$ to the original expression, we must add $32$ to the expression in order to balance the equation. \begin{align*} f(x) & = -2(x^2 - 8x) - 24\\ & = -2(x^2 - 8x + 16) - 24 + 32\\ & = -2(x - 4)^2 + 8 \end{align*} which tells us that the graph of $f(x)$ has vertex $(4, 8)$ and opens downwards. Note that the $y$-intercept is $$f(0) = -2(0 - 4)^2 + 8 = -2(16) + 8 = -32 + 8 = -24$$ as we would expect.

Check: We verify that the vertex form is equal to the original function. \begin{align*} f(x) & = -2(x - 4)^2 + 8\\ & = -2(x^2 - 8x + 16) + 8\\ & = -2x^2 + 16x - 32 + 8\\ & = -2x^2 + 16x - 24 \end{align*} Completing the square refers to the process of adding $k^2$ to an expression of the form $x^2 + 2kx$ to form the perfect square $x^2 + 2kx + k^2$. One purpose of completing the square is to transform a quadratic function into vertex form. What confused you is that it can also be used to solve for the $x$-intercepts. When we solve for the $x$-intercepts, we set $f(x) = 0$ and solve for $x$. It is here that we divide by $-2$ (or, more generally, the leading coefficient). Continuing from above yields \begin{align*} f(x) & = 0\\ -2(x - 4)^2 + 8 & = 0\\ (x - 4)^2 - 4 & = 0\\ (x - 4)^2 & = 4\\ x - 4 & = \pm 2\\ x & = 4 \pm 2 \end{align*} \begin{align*} x & = 4 + 2 & x & = 4 - 2\\ x & = 6 & x & = 2 \end{align*} You can verify that these solutions are correct by substituting them into the equation $f(x) = -2x^2 + 16x - 24$.