Question 174319
{{{y-x^2=-1}}} Start with the given equation.



{{{y=x^2-1}}} Add {{{x^2}}} to both sides.





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In order to graph {{{y=x^2-1}}}, we need to plot some points.



Let's find y when {{{x=-3}}}

(Note: you can start with any x-value):



{{{y=x^2-1}}} Start with the given equation.



{{{y=(-3)^2-1}}} Plug in {{{x=-3}}}.



{{{y=9-1}}}  Square {{{-3}}} to get {{{9}}}.



{{{y=8}}} Combine like terms.



So when {{{x=-3}}}, then {{{y=8}}}.



So we have the point *[Tex \LARGE \left(-3,8\right)].



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Let's find y when {{{x=-2}}}:



{{{y=x^2-1}}} Start with the given equation.



{{{y=(-2)^2-1}}} Plug in {{{x=-2}}}.



{{{y=4-1}}} Square {{{-2}}} to get {{{4}}}.



{{{y=3}}} Combine like terms.



So when {{{x=-2}}}, then {{{y=3}}}.



So we have the point *[Tex \LARGE \left(-2,3\right)].



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Let's find y when {{{x=-1}}}:



{{{y=x^2-1}}} Start with the given equation.



{{{y=(-1)^2-1}}} Plug in {{{x=-1}}}.



{{{y=1-1}}} Square {{{-1}}} to get {{{1}}}.



{{{y=0}}} Combine like terms.



So when {{{x=-1}}}, then {{{y=0}}}.



So we have the point *[Tex \LARGE \left(-1,0\right)].



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Let's find y when {{{x=0}}}:



{{{y=x^2-1}}} Start with the given equation.



{{{y=(0)^2-1}}} Plug in {{{x=0}}}.



{{{y=0-1}}} Square {{{0}}} to get {{{0}}}.



{{{y=-1}}} Combine like terms.



So when {{{x=0}}}, then {{{y=-1}}}.



So we have the point *[Tex \LARGE \left(0,-1\right)].



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Let's find y when {{{x=1}}}:



{{{y=x^2-1}}} Start with the given equation.



{{{y=(1)^2-1}}} Plug in {{{x=1}}}.



{{{y=1-1}}} Square {{{1}}} to get {{{1}}}.



{{{y=0}}} Combine like terms.



So when {{{x=1}}}, then {{{y=0}}}.



So we have the point *[Tex \LARGE \left(1,0\right)].



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Let's find y when {{{x=2}}}:



{{{y=x^2-1}}} Start with the given equation.



{{{y=(2)^2-1}}} Plug in {{{x=2}}}.



{{{y=4-1}}} Square {{{2}}} to get {{{4}}}.



{{{y=3}}} Combine like terms.



So when {{{x=2}}}, then {{{y=3}}}.



So we have the point *[Tex \LARGE \left(2,3\right)].



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Let's find y when {{{x=3}}}:



{{{y=x^2-1}}} Start with the given equation.



{{{y=(3)^2-1}}} Plug in {{{x=3}}}.



{{{y=9-1}}} Square {{{3}}} to get {{{9}}}.



{{{y=8}}} Combine like terms.



So when {{{x=3}}}, then {{{y=8}}}.



So we have the point *[Tex \LARGE \left(3,8\right)].



----------------------------------------------------



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Now let's make a table of the values we just found.



<h4>Table of Values:</h4><pre>

<TABLE border="1" width="100">
<TR><TD>x</TD><TD>y</TD></TR><tr><td>-3</td><td>8</td></tr>
<tr><td>-2</td><td>3</td></tr>
<tr><td>-1</td><td>0</td></tr>
<tr><td>0</td><td>-1</td></tr>
<tr><td>1</td><td>0</td></tr>
<tr><td>2</td><td>3</td></tr>
<tr><td>3</td><td>8</td></tr>
</TABLE>

</pre>

Now let's plot the points:



{{{ drawing(500, 500, -10, 10, -10, 10,
grid(1),
graph(500, 500, -10, 10, -10, 10, 0),
circle(-3,8,0.08),circle(-3,8,0.10),
circle(-2,3,0.08),circle(-2,3,0.10),
circle(-1,0,0.08),circle(-1,0,0.10),
circle(0,-1,0.08),circle(0,-1,0.10),
circle(1,0,0.08),circle(1,0,0.10),
circle(2,3,0.08),circle(2,3,0.10),
circle(3,8,0.08),circle(3,8,0.10)

)}}}


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<h4>Graph:</h4>

Now draw a curve through all of the points to graph {{{y=x^2-1}}}:



{{{ drawing(500, 500, -10, 10, -10, 10,
grid(1),
graph(500, 500, -10, 10, -10, 10, x^2-1),
circle(-3,8,0.08),circle(-3,8,0.10),
circle(-2,3,0.08),circle(-2,3,0.10),
circle(-1,0,0.08),circle(-1,0,0.10),
circle(0,-1,0.08),circle(0,-1,0.10),
circle(1,0,0.08),circle(1,0,0.10),
circle(2,3,0.08),circle(2,3,0.10),
circle(3,8,0.08),circle(3,8,0.10)

)}}} Graph of {{{y=x^2-1}}} (which is also the graph of {{{y-x^2=-1}}})



So the shape of {{{y-x^2=-1}}} is a parabola.