Finding the vertex of $y = x^2 - 6x + 1$. My solution doesn't match the book.

Question

I'm studying on how to find the vertex of a Parabola and got stuck with this question.

Determine the vertex of $y = x^2 - 6x + 1$

I found the vertex is $V=(3,-8)$ (see work below), but my workbook showed it as $V=(3,10)$. Unfortunately, it does not have a section where it shows how it got to that answer so I'm doubting if my understanding is incorrect. If I use online algebra calculators, it matches with mine but I just want to make sure that I understand how signs and the formula work.

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Formula I used to look for the Vertex: $$V=\left(-\frac{b}{2a} , \frac{4ac-b^2}{4a}\right)$$

My solution:

$$\begin{align} x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt] y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{-32}{4} = \frac{-8}{1}=-8 \end{align}$$

$$V=(3,-8)$$

What I think was done on the book:

$$\begin{align} x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt] y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{40}{4} = \frac{10}{1}=10 \end{align}$$

$$V=(3,10)$$

Is it correct that I should've added $4-36$ since they are both positive numbers or subtract it?

Any explanation is appreciated. Thanks!

Answer 1

Any general parabola can be written as $L_1^2=4AL_2$, if $L_1$ and $L_2$ are Eq. of non-parallel lines.

If $L_1$ and $L_2$ are perpendicular and $L_1$ and $L_2$ are normalized, life is simple. The length of latus rectum (LR) is 4A. The Eq. of axis of the parabola is $L_1=0$ Tangen at vertex is $L_1=0$. Vertex is found by solving $L_1=0$ and $L_2=0$. The Eq. of directrix is $L_2=-A$, Eq. of LR is $L_2=A$, the focus F come by solving $L_1=0$ & $L_2=A$.

So in your case $y=x^2-6x+1 \implies y+8= (x-3)^2$ Then $L_1=x-3, L_2=y+8, A=1/4$ The vertex is given by $L_1=0, L_2=0$ which is $x=3, y=-8$. the vertex is (3,-8)$.

Answer 2

$4-36$ is indeed $-32$. Whenever you have $a - b$ and $b > a$, you should interpret it as $a - b = -1 \cdot (b - a)$. So your calculation seems to be right. And yes, the vertex of the parabola is at $(3, -8)$, this can be verified by differentiating and setting $y' = 0$.

$y' = 2x - 6 = 0 \implies x = 3$ and then $y(x=3) = 3^2 - 6 \cdot 3 + 1 = 9 - 18 + 1 = -8$. Your method is correct, I just mentioned the differentiation method to confirm the answer.

Answer 3

Ah, well, both you and the book got the same $x$ coordinate for the vertex: $x=3$. So let's substitute $x=3$ into $y=x^2-6x+1$ and see the $y$ coordinate. It comes out as $y=-8$, so your solution is right.

(This shows there isn't much value in remembering the formula $y=\frac{4ac-b^2}{4a}$ - you can always just remember the formula for $x$ and then substitute $x$ into the original equality $y=ax^2+bx+c$ - the result will be the same.)

Answer 4

An easier way to do this is to take a derivative, solve for the minimum, this will give you, your x-value. The substituting that value back into the original equation we can get the y-value. For example here you have: y=(x^2)-6x+1 derivative: y'=(2x-6) Solving for x-value (solving for when y'=0) : 2x-6=0 -> x=3 <-X-value Now that we have the x value we can use this value in our original equation y=3^2 -6*3 +1 We find that y=-8 This gives us the coordinates for the vertex. This is probably the fastest approach here as taking the derivative of this function is pretty basic.. Hopefully this helps.