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

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

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Answer 1 · 2015-07-11T14:05:51

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

The axis of symmetry is the $y -$ $y-$ axis

Explanation:

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

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

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

$f (x) = x^{2}$ $f(x) = x^2$

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

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

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

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

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

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

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

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

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

$f (x) = - x^{2} = - {(x)}^{2}$ $f(x) = -x^2=-(x)^2$

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

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

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

So, for all values of $x$ $x$ 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 $10$ $10$ to our function:

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

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

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

where,
$m = s l o p e = \frac{r i s e}{r u n}$ $m=slope=(rise)/(run)$
we have our $x$ $x$ value
and $b$ $b$ 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 $f (x) = - x^{2} + 10$ $f(x)=-x^2+10$

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

Answer 2 · 2015-07-11T14:11:01

The vertex is at ( $0, 10$ $0,10$ ).
The axis of symmetry is $x = 0$ $x = 0$ .
The $y$ $y$ –intercept is at ( $0, 10$ $0,10$ ).
The $x$ $x$ -intercepts are at ( $- \sqrt{10}, 0$ $-sqrt10,0$ ) and ( $\sqrt{10}, 0$ $sqrt10,0$ ).

Explanation:

Your equation is

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

The standard form of the equation for a parabola is

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

So

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

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

The vertex form of a parabola is

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

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

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

So

$h = 0$ $h = 0$ and $k = 10$ $k = 10$

Vertex

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

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

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

Axis of symmetry

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

$y$ $y$ -intercept

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

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

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

$y = 10$ $y = 10$

The $y$ $y$ –intercept is at ( $0, 10$ $0,10$ ).

$x$ $x$ -intercepts

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

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

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

$x^{2} = 10$ $x^2 = 10$

$x = \pm \sqrt{10}$ $x = ±sqrt10$

The $x$ $x$ -intercepts are at ( $- \sqrt{10}, 0$ $-sqrt10,0$ ) and ( $\sqrt{10}, 0$ $sqrt10,0$ ).

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

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Explanation:

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How do find the vertex and axis of symmetry, and intercepts for a quadratic equation f(x) = -x^2 + 10f(x)=−x2+10f(x) = -x^2 + 10?

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Explanation:

Explanation:

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