# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation f(x) = -x^2 + 10?

Jul 11, 2015

The vertex is at the point $\left(0 , 10\right)$

The axis of symmetry is the $y -$axis

#### Explanation:

Let's look at the function $f \left(x\right) = - {x}^{2} + 10$

A function $f$ that depends on the variable $x$ is represented as $f \left(x\right)$

A function $f \left(x\right)$ being equal to a variable $x$ squared is represented as follows:

$f \left(x\right) = {x}^{2}$

Recall that ${x}^{2}$ when graphed looks like a "parabola" with it's "butt" resting on the origin, the point $\left(0 , 0\right)$.

graph{x^2 [-40, 40, -20, 20]}

Recall the coordinates for $x$ and $y$ are represented as $\left(x , y\right)$

So, $\left(0 , 0\right)$ means that:

$x = 0$ and $y = 0$ which gives us a point for our function $f \left(x\right) = {x}^{2}$

If we were to add in more points of interest we'd eventually draw our graph of ${x}^{2}$ above.

Such as $x = 2$, our function $f \left(x\right) = {x}^{2} = {2}^{2} = 4$
(our function is the curve drawn on the graph above)

...and so on with other values for $x$.

When our function is $- {x}^{2}$ this means:

$f \left(x\right) = - {x}^{2} = - {\left(x\right)}^{2}$

any value we put in for $x$ is that value squared and then take the negative of that squared value.

This just means that when we plug in a value for $x$, such as $2$, we get:

$f \left(x\right) = - {x}^{2} = - {\left(x\right)}^{2} = - {\left(2\right)}^{2} = - 4$

So, for all values of $x$ we plug into our function our curve on our graph curves downward:

graph{-x^2 [-40, 40, -20, 20]}

Lastly, if we were to add $10$ to our function:

$f \left(x\right) = - {x}^{2} + 10$

this means that our curve's "butt" gets moved up the y-axis by $10$, as seen by our graph below of the function $f \left(x\right) = - {x}^{2} + 10$

Recall the slope-intercept form: $y = m x + b$

where,
$m = s l o p e = \frac{r i s e}{r u n}$
we have our $x$ value
and $b$ is the y-intercept (where the function crosses the y-axis)

graph{-x^2+10 [-40, 40, -20, 20]}

Now, I've been saying "butt" of the function, but that "butt" is our vertex of our function $f \left(x\right) = - {x}^{2} + 10$

So, the vertex is at $y = 10$ and $x = 0$ or $\left(0 , 10\right)$ and as we can tell by looking at our graph that the axis of symmetry is the $y -$axis because if we were to cut our graph in two at $\left(0 , 10\right)$ we would see that we would have a mirror image of our curve on both sides of the $y -$axis

Jul 11, 2015

The vertex is at ($0 , 10$).
The axis of symmetry is $x = 0$.
The $y$–intercept is at ($0 , 10$).
The $x$-intercepts are at ($- \sqrt{10} , 0$) and ($\sqrt{10} , 0$).

#### Explanation:

$f \left(x\right) = - {x}^{2} + 10$

The standard form of the equation for a parabola is

$y = a {x}^{2} + b x + c$

So

$a = - 1$, $b = 0$, and $c = 10$.

Let's put your equation into "vertex form".

The vertex form of a parabola is

$y = a {\left(y - h\right)}^{2} + k$

$h = - \frac{b}{2 a} = - \frac{0}{2 \left(- 1\right)} = 0$

$k = f \left(h\right) = - {0}^{2} + 10 = 10$

So

$h = 0$ and $k = 10$

Vertex

Since $a < 0$, the parabola opens downwards.

The vertex is at ($0 , k$), and $k = 10$,

So the vertex is at ($0 , 10$).

Axis of symmetry

The axis of symmetry is $x = h$ or $x = 0$.

$y$-intercept

Set $x = 0$ and solve for $y$.

$y = - {x}^{2} + 10 = 0$

$y = - {0}^{2} + 10$

$y = 10$

The $y$–intercept is at ($0 , 10$).

$x$-intercepts

Set $y = 0$ and solve for $x$.

$y = - {x}^{2} + 10$

$0 = - {x}^{2} + 10$

${x}^{2} = 10$

x = ±sqrt10

The $x$-intercepts are at ($- \sqrt{10} , 0$) and ($\sqrt{10} , 0$).