How do find the vertex and axis of symmetry, and intercepts for a quadratic equation f(x)=−x2+10?

2 Answers
Erudite
Jul 11, 2015

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)=−x2+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)=x2

Recall that x2 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)=x2

If we were to add in more points of interest we'd eventually draw our graph of x2 above.

Such as x=2, our function f(x)=x2=22=4
(our function is the curve drawn on the graph above)

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

When our function is −x2 this means:

f(x)=−x2=−(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)=−x2=−(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)=−x2+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)=−x2+10

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

where,
m=slope=riserun
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)=−x2+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

Ernest Z.
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 (−√10,0) and (√10,0).

Explanation:

Your equation is

f(x)=−x2+10

The standard form of the equation for a parabola is

y=ax2+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=−b2a=−02(−1)=0

k=f(h)=−02+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=−x2+10=0

y=−02+10

y=10

The y–intercept is at (0,10).

x-intercepts

Set y=0 and solve for x.

y=−x2+10

0=−x2+10

x2=10

x=±√10

The x-intercepts are at (−√10,0) and (√10,0).

enter image source here

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