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

2 Answers
Jul 27, 2018

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Please read the explanation.

Explanation:

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A quadratic equation is of the form:

color(red)(ax^2+bx+c=0

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

Set color(blue)(y=0

We have the quadratic equation:

color(blue)(x^2-9=0

Using the algebraic identity:

color(green)(a^2-b^2 -= (a+b)(a-b)

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

x^2-3^2=0

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

rArr (x+3)=0, (x-3)=0

rArr x= -3, x=3

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

So, we have two x-intercepts:

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

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

rArr y=(0)^2-9=0

rArr y = -9

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

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

enter image source here

Hope it helps.

Jul 27, 2018

Refer to the explanation.

Explanation:

Given:

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

y=ax^2+bx+c,

where:

a=1, b=0, and 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) of the parabola

Since 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=(-b)/(2a)

x=0/2

x=0

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

y=0^2-9

y=-9

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

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

X-intercepts: values for x when y=0

Substitute 0 for y and solve for x.

0=x^2-9

Switch sides.

x^2-9=0

Factor x^-9

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

Set each binomial equal to 0 and solve.

x+3=0

x=-3

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

x-3=0

x=3

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

For additional points, choose values for x and solve for 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]}