Write the quadratic function in the form $g(x)= a(x-h)^2 + k$

Question

Write the quadratic function in the form $g(x)= a(x-h)^2 + k$

Then, give the vertex of its graph.

$g(x)= 2x^2-16x+35$

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I tried, but still lost:

$(2x^2-16x)+35$

$2(x^2-4^2-16)+35$

$2(x^2-4^2-32+35$

Answer 1

$2x^2-16x+35$

$=2(x^2-8x)+35$

$=2(x^2-8x+16)-2\cdot 16+35$

$=2(x-4)^2+3$

There you go. Completing the square.

The vertex is easy from this. It's just x=4 since squares are always nonnegative so to make the square part zero you let x=4. (to be more specific the vertex is (4, 3))

Answer 2

I'm not too sure what you did, one of your brackets is missing, but I believe that your error lies in $2(x^2-4^2-16)+35$.

When you complete the square for $x^2+bx$, you take half of $b$ and then square that. However, starting with $x^2+16x$, you subtracted $16$, and then "turned" the $8x$ into a $4^2$. Not possible!

Instead, we're supposed to have$$\begin{align*}2x^2-16x+35 & =2(x^2-8x)+35\\ & =2\left\{x^2-8x+\left(\dfrac {8}2\right)^2-\left(\dfrac 82\right)^2\right\}+35\\ & =2\left(x^2-8x+16\right)+35-32\\ & =2(x-4)^2+3\end{align*}$$ Thus, the vertex of $g(x)$ is $(4,3)$.

Answer 3

You have few mistakes.

It should be

$2(x^2 + 4^2 - 8x) + 35 - 2 \cdot 4^2$

= $2(x - 4)^2 + 35 - 32$

= $2(x - 4)^2 + 3$

Answer 4

It may be prudent to write out all of the steps without any shortcuts. $$\begin{align} g(x)&=2x^2 - 16x + 35\\ g(x)&=2(x^2 - 8x) + 35\\ g(x)&=2(x^2 - 8x) + 35\\ g(x)&=2(x^2 - 4x -4x) + 35\\ g(x)&=2(x^2 - 4x -4x + 0) + 35\\ g(x)&=2(x^2 - 4x -4x + 16 - 16) + 35\\ g(x)&=2(x^2 - 4x -4x + 16) -32 + 35\\ g(x)&=2(x(x - 4) -4(x - 4)) +3\\ g(x)&=2(x - 4)(x - 4) +3\\ g(x)&=2(x - 4)^2 +3\\ \end{align}$$

You know that the vertex will occur when $x-h = 0$. Maybe draw a table of values, or draw a graph, or both to determine the coordinates of the vertex.