What is the vertex of  y= x^2 -9 - 8x?

Feb 3, 2016

The vertex is $\left(4 , - 25\right)$.

Explanation:

First place the equation in standard form.

$y = {x}^{2} - 8 x - 9$

This is a quadratic equation in standard form, $a {x}^{2} + b x + c$, where $a = 1 , b = - 8 , c = - 9$.

The vertex is the maximum or minimum point of a parabola. In this case, since $a > 0$, the parabola opens upward and the vertex is the minimum point.

To find the vertex of a parabola in standard form, first find the axis of symmetry, which will give us $x$. The axis of symmetry is the imaginary line that divides a parabola into two equal halves. Once we have $x$, we can substitute it into the equation and solve for $y$, giving us the $y$ value for the vertex.

Axis of Symmetry

$x = \frac{- b}{2 a}$

Substitute the values for $a$ and $b$ into the equation.

$x = \frac{- \left(- 8\right)}{2 \cdot 1}$

Simplify.

$x = \frac{8}{2}$

$x = 4$

Determine the value for $y$.

Substitute $4$ for $x$ into the equation.

$y = {4}^{2} - \left(8 \cdot 4\right) - 9$

Simplify.

$y = 16 - 32 - 9$

Simplify.

$y = - 25$

Vertex = $\left(x , y\right)$=$\left(4 , - 25\right)$.

graph{y=x^2-8x-9 [-10.21, 7.01, -26.63, -18.02]}

Feb 3, 2016

$\left(4 , - 25\right)$

Explanation:

We are given $y = {x}^{2} - 9 - 8 x$.

First I want to get this into standard form This is easy, we just need to reorder it to fit the $a {x}^{2} + b x + c$ form.

Now we have ${x}^{2} - 8 x - 9$. The easiest way to get a standard form into vertex form is by completing the square. The process of completing the square is making ${x}^{2} - 8 x + \left(b l a n k\right)$ a perfect square. We just need to find the value that completes that. First we take the middle term, $- 8 x$, and divide it by 2 (so $- \frac{8}{2}$, which is $- 4$). Then we square that answer, ${\left(- 4\right)}^{2}$, which is $16$.

Now we plug in $16$ in to the equation to make a perfect square, right?

Well, let's take a look at that: ${x}^{2} - 8 x + 16 - 9 = y$. Now, look again. We can't just add a random number on one side of an equation and not add it on the other side. What we do to one side we must do to the other. So now we have ${x}^{2} - 8 x + 16 - 9 = y + 16$.

After we did all this work, let's make ${x}^{2} - 8 x + 16$ into a perfect square, which looks like this ${\left(x - 4\right)}^{2}$. Replace ${x}^{2} - 8 x + 16$ with it and we have ${\left(x - 4\right)}^{2} - 9 = y + 16$. Now I don't know about you, but I liked having $y$ isolated, so let's get it alone by subtracting $16$ on both sides.

Now we've got ${\left(x - 4\right)}^{2} - 9 - 16 = y$, which we can simplify to ${\left(x - 4\right)}^{2} - 25 = y$.

Now this is in vertex form, and once we have that it's very quick to find the vertex. This is vertex form,$y = a {\left(x - \textcolor{red}{h}\right)}^{2} \textcolor{b l u e}{+ k}$, and the vertex from that is $\left(\textcolor{red}{h , \textcolor{b l u e}{k}}\right)$.

In the case of our equation we have $y = {\left(x - \textcolor{red}{4}\right)}^{2} \textcolor{b l u e}{- 25}$, or $\left(\textcolor{red}{4} , \textcolor{b l u e}{- 25}\right)$.

PLEASE NOTE that $\left(\textcolor{red}{h} , k\right)$ is the opposite of what it was in the equation!
example: $y = {\left(x + 3\right)}^{2} + 3$, vertex is $\left(\textcolor{red}{-} 3 , 3\right)$.

So, the vertex is $\left(4 , - 25\right)$, and we can check this by graphing the equation and finding the vertex, which is the highest or lowest point on the parabola.
graph{x^2-8x-9}

Looks like we got it right!! Nice job!