How do you graph $y = x^{2} - 9$ $y=x^2-9$ ?

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How do you graph $y = x^{2} - 9$ $y=x^2-9$ ?

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Answer 1 · 2018-07-27T01:44:06

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Please read the explanation.

Explanation:

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A quadratic equation is of the form:

$a x^{2} + b x + c = 0$ $color(red)(ax^2+bx+c=0$

We have : $y = f (x) = x^{2} - 9$ $color(red)(y=f(x)=x^2-9$

Set $y = 0$ $color(blue)(y=0$

We have the quadratic equation:

$x^{2} - 9 = 0$ $color(blue)(x^2-9=0$

Using the algebraic identity:

$a^{2} - b^{2} \equiv (a + b) (a - b)$ $color(green)(a^2-b^2 -= (a+b)(a-b)$

We can rewrite $x^{2} - 9 = 0$ $color(blue)(x^2-9=0$ as

$x^{2} - 3^{2} = 0$ $x^2-3^2=0$

$\Rightarrow (x + 3) (x - 3) = 0$ $rArr (x+3)(x-3)=0$

$\Rightarrow (x + 3) = 0, (x - 3) = 0$ $rArr (x+3)=0, (x-3)=0$

$\Rightarrow x = - 3, x = 3$ $rArr x= -3, x=3$

Hence, there are two solutions for $x$ $color(red)(x$

So, we have two x-intercepts:

$(- 3, 0) and (3, 0)$ $color(red)((-3,0) and (3,0)$

To find the y-intercept, set $x = 0$ $color(blue)(x=0$

$\Rightarrow y = {(0)}^{2} - 9 = 0$ $rArr y=(0)^2-9=0$

$\Rightarrow y = - 9$ $rArr y = -9$

Hence, the y-intercept: $(0, - 9)$ $color(red)((0,-9)$

The graph of $y = f (x) = x^{2} - 9$ $color(red)(y=f(x)=x^2-9$ is given below:

enter image source here

Hope it helps.

Answer 2 · 2018-07-27T01:59:18

Refer to the explanation.

Explanation:

Given:

$y = x^{2} - 9$ $y=x^2-9$ is a quadratic equation in standard form:

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

where:

$a = 1$ $a=1$ , $b = 0$ $b=0$ , and $c = - 9$ $c=-9$

To graph a quadratic equation in standard form, you need the vertex, y-intercept, x-intercepts (if real), and one or two additional points.

Vertex: maximum or minimum point $(x, y)$ $(x,y)$ of the parabola

Since $a > 0$ $a>0$ , the vertex is the minimum point and the parabola opens upward.

The x-coordinate of the vertex is determined using the formula for the axis of symmetry:

$x = \frac{- b}{2 a}$ $x=(-b)/(2a)$

$x = \frac{0}{2}$ $x=0/2$

$x = 0$ $x=0$

To find the y-coordinate of the vertex, substitute $0$ $0$ for $x$ $x$ and solve for $y$ $y$ .

$y = 0^{2} - 9$ $y=0^2-9$

$y = - 9$ $y=-9$

The vertex is $(0, - 9)$ $(0,-9)$ Plot this point.

In this case, the vertex is also the y-intercept, which is the value of $y$ $y$ when $x = 0$ $x=0$ .

X-intercepts: values for $x$ $x$ when $y = 0$ $y=0$

Substitute $0$ $0$ for $y$ $y$ and solve for $x$ $x$ .

$0 = x^{2} - 9$ $0=x^2-9$

Switch sides.

$x^{2} - 9 = 0$ $x^2-9=0$

Factor $x^{- 9}$ $x^-9$

$(x + 3) (x - 3) = 0$ $(x+3)(x-3)=0$

Set each binomial equal to $0$ $0$ and solve.

$x + 3 = 0$ $x+3=0$

$x = - 3$ $x=-3$

Point: $(- 3, 0)$ $(-3,0)$ Plot this point. $\leftarrow$ $larr$ first x-intercept

$x - 3 = 0$ $x-3=0$

$x = 3$ $x=3$

Point: $(3, 0)$ $(3,0)$ Plot this point. $\leftarrow$ $larr$ second x-intercept

For additional points, choose values for $x$ $x$ and solve for $y$ $y$ .

Plot all the points and sketch a parabola through the points. Do not connect the dots.

graph{y=x^2-9 [-11.13, 11.37, -9.885, 1.365]}

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