SOLUTION: Rewrite each equation in vertex form. Then sketch the graph. 1. a. y=x^2+4x-6 b.y=4x^2+4x+1 c.y=-3x^2+3x-1 d. y=-2x^2+4x+3 e.y=6x^2-12x+1 I dont kn

Algebra ->  Linear-equations -> SOLUTION: Rewrite each equation in vertex form. Then sketch the graph. 1. a. y=x^2+4x-6 b.y=4x^2+4x+1 c.y=-3x^2+3x-1 d. y=-2x^2+4x+3 e.y=6x^2-12x+1 I dont kn      Log On


   

Question 83111: Rewrite each equation in vertex form. Then sketch the graph.
1. a. y=x^2+4x-6
b.y=4x^2+4x+1
c.y=-3x^2+3x-1
d. y=-2x^2+4x+3

e.y=6x^2-12x+1
I dont know how to do these they are really confusing
HELP!!!Please and Thanks
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%2B4+x-6 Start with the given equation



y%2B6=1+x%5E2%2B4+x Add 6 to both sides



y%2B6=1%28x%5E2%2B4x%29 Factor out the leading coefficient 1



Take half of the x coefficient 4 to get 2 (ie %281%2F2%29%284%29=2).


Now square 2 to get 4 (ie %282%29%5E2=%282%29%282%29=4)





y%2B6=1%28x%5E2%2B4x%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%2B6=1%28%28x%2B2%29%5E2-4%29 Now factor x%5E2%2B4x%2B4 to get %28x%2B2%29%5E2



y%2B6=1%28x%2B2%29%5E2-1%284%29 Distribute



y%2B6=1%28x%2B2%29%5E2-4 Multiply



y=1%28x%2B2%29%5E2-4-6 Now add %2B6 to both sides to isolate y



y=1%28x%2B2%29%5E2-10 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-2, and k=-10. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2%2B4x-6 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B4x-6%29 Graph of y=1x%5E2%2B4x-6. Notice how the vertex is (-2,-10).



Notice if we graph the final equation y=1%28x%2B2%29%5E2-10 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x%2B2%29%5E2-10%29 Graph of y=1%28x%2B2%29%5E2-10. Notice how the vertex is also (-2,-10).



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=4+x%5E2%2B4+x%2B1 Start with the given equation



y-1=4+x%5E2%2B4+x Subtract 1 from both sides



y-1=4%28x%5E2%2B1x%29 Factor out the leading coefficient 4



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


Now square 1%2F2 to get 1%2F4 (ie %281%2F2%29%5E2=%281%2F2%29%281%2F2%29=1%2F4)





y-1=4%28x%5E2%2B1x%2B1%2F4-1%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1%2F4 does not change the equation




y-1=4%28%28x%2B1%2F2%29%5E2-1%2F4%29 Now factor x%5E2%2B1x%2B1%2F4 to get %28x%2B1%2F2%29%5E2



y-1=4%28x%2B1%2F2%29%5E2-4%281%2F4%29 Distribute



y-1=4%28x%2B1%2F2%29%5E2-1 Multiply



y=4%28x%2B1%2F2%29%5E2-1%2B1 Now add 1 to both sides to isolate y



y=4%28x%2B1%2F2%29%5E2%2B0 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=4, h=-1%2F2, 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=4x%5E2%2B4x%2B1 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C4x%5E2%2B4x%2B1%29 Graph of y=4x%5E2%2B4x%2B1. Notice how the vertex is (-1%2F2,0).



Notice if we graph the final equation y=4%28x%2B1%2F2%29%5E2%2B0 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C4%28x%2B1%2F2%29%5E2%2B0%29 Graph of y=4%28x%2B1%2F2%29%5E2%2B0. Notice how the vertex is also (-1%2F2,0).



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=-3+x%5E2%2B3+x-1 Start with the given equation



y%2B1=-3+x%5E2%2B3+x Add 1 to both sides



y%2B1=-3%28x%5E2-1x%29 Factor out the leading coefficient -3



Take half of the x coefficient -1 to get -1%2F2 (ie %281%2F2%29%28-1%29=-1%2F2).


Now square -1%2F2 to get 1%2F4 (ie %28-1%2F2%29%5E2=%28-1%2F2%29%28-1%2F2%29=1%2F4)





y%2B1=-3%28x%5E2-1x%2B1%2F4-1%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1%2F4 does not change the equation




y%2B1=-3%28%28x-1%2F2%29%5E2-1%2F4%29 Now factor x%5E2-1x%2B1%2F4 to get %28x-1%2F2%29%5E2



y%2B1=-3%28x-1%2F2%29%5E2%2B3%281%2F4%29 Distribute



y%2B1=-3%28x-1%2F2%29%5E2%2B3%2F4 Multiply



y=-3%28x-1%2F2%29%5E2%2B3%2F4-1 Now add %2B1 to both sides to isolate y



y=-3%28x-1%2F2%29%5E2-1%2F4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=-3, h=1%2F2, and k=-1%2F4. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=-3x%5E2%2B3x-1 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-3x%5E2%2B3x-1%29 Graph of y=-3x%5E2%2B3x-1. Notice how the vertex is (1%2F2,-1%2F4).



Notice if we graph the final equation y=-3%28x-1%2F2%29%5E2-1%2F4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-3%28x-1%2F2%29%5E2-1%2F4%29 Graph of y=-3%28x-1%2F2%29%5E2-1%2F4. Notice how the vertex is also (1%2F2,-1%2F4).



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=-2+x%5E2%2B4+x%2B3 Start with the given equation



y-3=-2+x%5E2%2B4+x Subtract 3 from both sides



y-3=-2%28x%5E2-2x%29 Factor out the leading coefficient -2



Take half of the x coefficient -2 to get -1 (ie %281%2F2%29%28-2%29=-1).


Now square -1 to get 1 (ie %28-1%29%5E2=%28-1%29%28-1%29=1)





y-3=-2%28x%5E2-2x%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-3=-2%28%28x-1%29%5E2-1%29 Now factor x%5E2-2x%2B1 to get %28x-1%29%5E2



y-3=-2%28x-1%29%5E2%2B2%281%29 Distribute



y-3=-2%28x-1%29%5E2%2B2 Multiply



y=-2%28x-1%29%5E2%2B2%2B3 Now add 3 to both sides to isolate y



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




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=-2, h=1, 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=-2x%5E2%2B4x%2B3 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-2x%5E2%2B4x%2B3%29 Graph of y=-2x%5E2%2B4x%2B3. Notice how the vertex is (1,5).



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


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



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=6+x%5E2-12+x%2B1 Start with the given equation



y-1=6+x%5E2-12+x Subtract 1 from both sides



y-1=6%28x%5E2-2x%29 Factor out the leading coefficient 6



Take half of the x coefficient -2 to get -1 (ie %281%2F2%29%28-2%29=-1).


Now square -1 to get 1 (ie %28-1%29%5E2=%28-1%29%28-1%29=1)





y-1=6%28x%5E2-2x%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=6%28%28x-1%29%5E2-1%29 Now factor x%5E2-2x%2B1 to get %28x-1%29%5E2



y-1=6%28x-1%29%5E2-6%281%29 Distribute



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



y=6%28x-1%29%5E2-6%2B1 Now add 1 to both sides to isolate y



y=6%28x-1%29%5E2-5 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=6, h=1, 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=6x%5E2-12x%2B1 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C6x%5E2-12x%2B1%29 Graph of y=6x%5E2-12x%2B1. Notice how the vertex is (1,-5).



Notice if we graph the final equation y=6%28x-1%29%5E2-5 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C6%28x-1%29%5E2-5%29 Graph of y=6%28x-1%29%5E2-5. Notice how the vertex is also (1,-5).



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