Question

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

Answer 1

The vertex is at the point `(0,10)`

The axis of symmetry is the `y-`axis

Explanation:

Let's look at the function `f(x) = -x^2+10`

A function `f` that depends on the variable `x` is represented as `f(x)`

A function `f(x)` being equal to a variable `x` squared is represented as follows:

`f(x) = x^2`

Recall that `x^2` when graphed looks like a "parabola" with it's "butt" resting on the origin, the point `(0,0)`.

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

Recall the coordinates for `x` and `y` are represented as `(x,y)`

So, `(0,0)` means that:

`x=0` and `y=0` which gives us a point for our function `f(x)=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(x)=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(x) = -x^2=-(x)^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(x) = -x^2=-(x)^2=-(2)^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(x)=-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(x)=-x^2+10`

Recall the slope-intercept form: `y=mx+b`

where,
`m=slope=(rise)/(run)`
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(x)=-x^2+10`

So, the vertex is at `y=10` and `x=0` or `(0,10)` 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 `(0,10)` we would see that we would have a mirror image of our curve on both sides of the `y-`axis

Answer 2

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 (`-sqrt10,0`) and (`sqrt10,0`).

Explanation:

Your equation is

`f(x) = -x^2+10`

The standard form of the equation for a parabola is

`y = ax^2 + bx + 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(y-h)^2 +k`

`h = -b/(2a) = -0/(2(-1)) = 0`

`k = f(h) = -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 (`-sqrt10,0`) and (`sqrt10,0`).