SOLUTION: Rewrite each equation in vertex form.Then Sketch the graph 7. a. y=x^2-6x+6 b. y=2x62+4x-5 c.y=x^2+2x+1 d.y=x^2+5x+5/4 e.y=-2x^2+8x-9

Algebra ->  Linear-equations -> SOLUTION: Rewrite each equation in vertex form.Then Sketch the graph 7. a. y=x^2-6x+6 b. y=2x62+4x-5 c.y=x^2+2x+1 d.y=x^2+5x+5/4 e.y=-2x^2+8x-9      Log On


   

Question 83112: Rewrite each equation in vertex form.Then Sketch the graph
7. a. y=x^2-6x+6
b. y=2x62+4x-5
c.y=x^2+2x+1
d.y=x^2+5x+5/4
e.y=-2x^2+8x-9
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2-6+x%2B6 Start with the given equation



y-6=1+x%5E2-6+x Subtract 6 from both sides



y-6=1%28x%5E2-6x%29 Factor out the leading coefficient 1



Take half of the x coefficient -6 to get -3 (ie %281%2F2%29%28-6%29=-3).


Now square -3 to get 9 (ie %28-3%29%5E2=%28-3%29%28-3%29=9)





y-6=1%28x%5E2-6x%2B9-9%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 9 does not change the equation




y-6=1%28%28x-3%29%5E2-9%29 Now factor x%5E2-6x%2B9 to get %28x-3%29%5E2



y-6=1%28x-3%29%5E2-1%289%29 Distribute



y-6=1%28x-3%29%5E2-9 Multiply



y=1%28x-3%29%5E2-9%2B6 Now add 6 to both sides to isolate y



y=1%28x-3%29%5E2-3 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=3, and k=-3. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-6x%2B6 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-6x%2B6%29 Graph of y=1x%5E2-6x%2B6. Notice how the vertex is (3,-3).



Notice if we graph the final equation y=1%28x-3%29%5E2-3 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-3%29%5E2-3%29 Graph of y=1%28x-3%29%5E2-3. Notice how the vertex is also (3,-3).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.






b)
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=2+x%5E2%2B4+x-5 Start with the given equation



y%2B5=2+x%5E2%2B4+x Add 5 to both sides



y%2B5=2%28x%5E2%2B2x%29 Factor out the leading coefficient 2



Take half of the x coefficient 2 to get 1 (ie %281%2F2%29%282%29=1).


Now square 1 to get 1 (ie %281%29%5E2=%281%29%281%29=1)





y%2B5=2%28x%5E2%2B2x%2B1-1%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1 does not change the equation




y%2B5=2%28%28x%2B1%29%5E2-1%29 Now factor x%5E2%2B2x%2B1 to get %28x%2B1%29%5E2



y%2B5=2%28x%2B1%29%5E2-2%281%29 Distribute



y%2B5=2%28x%2B1%29%5E2-2 Multiply



y=2%28x%2B1%29%5E2-2-5 Now add %2B5 to both sides to isolate y



y=2%28x%2B1%29%5E2-7 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=2, h=-1, and k=-7. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=2x%5E2%2B4x-5 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C2x%5E2%2B4x-5%29 Graph of y=2x%5E2%2B4x-5. Notice how the vertex is (-1,-7).



Notice if we graph the final equation y=2%28x%2B1%29%5E2-7 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C2%28x%2B1%29%5E2-7%29 Graph of y=2%28x%2B1%29%5E2-7. Notice how the vertex is also (-1,-7).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.







c)
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2%2B2+x%2B1 Start with the given equation



y-1=1+x%5E2%2B2+x Subtract 1 from both sides



y-1=1%28x%5E2%2B2x%29 Factor out the leading coefficient 1



Take half of the x coefficient 2 to get 1 (ie %281%2F2%29%282%29=1).


Now square 1 to get 1 (ie %281%29%5E2=%281%29%281%29=1)





y-1=1%28x%5E2%2B2x%2B1-1%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1 does not change the equation




y-1=1%28%28x%2B1%29%5E2-1%29 Now factor x%5E2%2B2x%2B1 to get %28x%2B1%29%5E2



y-1=1%28x%2B1%29%5E2-1%281%29 Distribute



y-1=1%28x%2B1%29%5E2-1 Multiply



y=1%28x%2B1%29%5E2-1%2B1 Now add 1 to both sides to isolate y



y=1%28x%2B1%29%5E2%2B0 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-1, and k=0. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2%2B2x%2B1 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B2x%2B1%29 Graph of y=1x%5E2%2B2x%2B1. Notice how the vertex is (-1,0).



Notice if we graph the final equation y=1%28x%2B1%29%5E2%2B0 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x%2B1%29%5E2%2B0%29 Graph of y=1%28x%2B1%29%5E2%2B0. Notice how the vertex is also (-1,0).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.







d)
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2%2B5+x%2B5%2F4 Start with the given equation



y-5%2F4=1+x%5E2%2B5+x Subtract 5%2F4 from both sides



y-5%2F4=1%28x%5E2%2B5x%29 Factor out the leading coefficient 1



Take half of the x coefficient 5 to get 5%2F2 (ie %281%2F2%29%285%29=5%2F2).


Now square 5%2F2 to get 25%2F4 (ie %285%2F2%29%5E2=%285%2F2%29%285%2F2%29=25%2F4)





y-5%2F4=1%28x%5E2%2B5x%2B25%2F4-25%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 25%2F4 does not change the equation




y-5%2F4=1%28%28x%2B5%2F2%29%5E2-25%2F4%29 Now factor x%5E2%2B5x%2B25%2F4 to get %28x%2B5%2F2%29%5E2



y-5%2F4=1%28x%2B5%2F2%29%5E2-1%2825%2F4%29 Distribute



y-5%2F4=1%28x%2B5%2F2%29%5E2-25%2F4 Multiply



y=1%28x%2B5%2F2%29%5E2-25%2F4%2B5%2F4 Now add 5%2F4 to both sides to isolate y



y=1%28x%2B5%2F2%29%5E2-5 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-5%2F2, and k=-5. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2%2B5x%2B5%2F4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B5x%2B5%2F4%29 Graph of y=1x%5E2%2B5x%2B5%2F4. Notice how the vertex is (-5%2F2,-5).



Notice if we graph the final equation y=1%28x%2B5%2F2%29%5E2-5 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x%2B5%2F2%29%5E2-5%29 Graph of y=1%28x%2B5%2F2%29%5E2-5. Notice how the vertex is also (-5%2F2,-5).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.








e)
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=-2+x%5E2%2B8+x-9 Start with the given equation



y%2B9=-2+x%5E2%2B8+x Add 9 to both sides



y%2B9=-2%28x%5E2-4x%29 Factor out the leading coefficient -2



Take half of the x coefficient -4 to get -2 (ie %281%2F2%29%28-4%29=-2).


Now square -2 to get 4 (ie %28-2%29%5E2=%28-2%29%28-2%29=4)





y%2B9=-2%28x%5E2-4x%2B4-4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 4 does not change the equation




y%2B9=-2%28%28x-2%29%5E2-4%29 Now factor x%5E2-4x%2B4 to get %28x-2%29%5E2



y%2B9=-2%28x-2%29%5E2%2B2%284%29 Distribute



y%2B9=-2%28x-2%29%5E2%2B8 Multiply



y=-2%28x-2%29%5E2%2B8-9 Now add %2B9 to both sides to isolate y



y=-2%28x-2%29%5E2-1 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=-2, h=2, and k=-1. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=-2x%5E2%2B8x-9 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-2x%5E2%2B8x-9%29 Graph of y=-2x%5E2%2B8x-9. Notice how the vertex is (2,-1).



Notice if we graph the final equation y=-2%28x-2%29%5E2-1 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-2%28x-2%29%5E2-1%29 Graph of y=-2%28x-2%29%5E2-1. Notice how the vertex is also (2,-1).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.