SOLUTION: Write the equation of the axis of symmetry and find the coordinates of the vertex of the graph of each equation. y=3xsqaured+4 y=3xsqaured+6x-17 y=3(x+1)sqaured-20 y=xs

Algebra ->  Graphs -> SOLUTION: Write the equation of the axis of symmetry and find the coordinates of the vertex of the graph of each equation. y=3xsqaured+4 y=3xsqaured+6x-17 y=3(x+1)sqaured-20 y=xs      Log On


   

Question 84539: Write the equation of the axis of symmetry and find the coordinates of the vertex of the graph of each equation.
y=3xsqaured+4
y=3xsqaured+6x-17
y=3(x+1)sqaured-20
y=xsqaured+2x
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
1.
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=3+x%5E2%2B0+x%2B4 Start with the given equation



y-4=3+x%5E2%2B0+x Subtract 4 from both sides



y-4=3%28x%5E2%2B%280%29x%29 Factor out the leading coefficient 3



Take half of the x coefficient 0 to get 0 (ie %281%2F2%29%280%29=0).


Now square 0 to get 0 (ie %280%29%5E2=%280%29%280%29=0)





y-4=3%28x%5E2%2B%280%29x%2B0-0%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 0 does not change the equation




y-4=3%28%28x%2B0%29%5E2-0%29 Now factor x%5E2%2B%280%29x%2B0 to get %28x%2B0%29%5E2



y-4=3%28x%2B0%29%5E2-3%280%29 Distribute



y-4=3%28x%2B0%29%5E2-0 Multiply



y=3%28x%2B0%29%5E2-0%2B4 Now add 4 to both sides to isolate y



y=3%28x%2B0%29%5E2%2B4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=3, h=-0, and k=4. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=3x%5E2%2B0x%2B4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C3x%5E2%2B0x%2B4%29 Graph of y=3x%5E2%2B0x%2B4. Notice how the vertex is (-0,4).



Notice if we graph the final equation y=3%28x%2B0%29%5E2%2B4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C3%28x%2B0%29%5E2%2B4%29 Graph of y=3%28x%2B0%29%5E2%2B4. Notice how the vertex is also (-0,4).



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






2.
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=3+x%5E2%2B6+x-17 Start with the given equation



y%2B17=3+x%5E2%2B6+x Add 17 to both sides



y%2B17=3%28x%5E2%2B2x%29 Factor out the leading coefficient 3



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%2B17=3%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%2B17=3%28%28x%2B1%29%5E2-1%29 Now factor x%5E2%2B2x%2B1 to get %28x%2B1%29%5E2



y%2B17=3%28x%2B1%29%5E2-3%281%29 Distribute



y%2B17=3%28x%2B1%29%5E2-3 Multiply



y=3%28x%2B1%29%5E2-3-17 Now add %2B17 to both sides to isolate y



y=3%28x%2B1%29%5E2-20 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=3, h=-1, and k=-20. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=3x%5E2%2B6x-17 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C3x%5E2%2B6x-17%29 Graph of y=3x%5E2%2B6x-17. Notice how the vertex is (-1,-20).



Notice if we graph the final equation y=3%28x%2B1%29%5E2-20 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C3%28x%2B1%29%5E2-20%29 Graph of y=3%28x%2B1%29%5E2-20. Notice how the vertex is also (-1,-20).



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






3.
Since the equation y=3%28x%2B1%29%5E2-20 is already in vertex form, the equation of the axis of symmetry is x=-1 and the vertex is (-1,-20). Remember, any equation in vertex form y=a%28x-h%29%5E2%2Bk has an axis of symmetry of x=h and a vertex of (h,k)


4.
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2%2B2+x%2B0 Start with the given equation



y-0=1+x%5E2%2B2+x Subtract 0 from both sides



y-0=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-0=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-0=1%28%28x%2B1%29%5E2-1%29 Now factor x%5E2%2B2x%2B1 to get %28x%2B1%29%5E2



y-0=1%28x%2B1%29%5E2-1%281%29 Distribute



y-0=1%28x%2B1%29%5E2-1 Multiply



y=1%28x%2B1%29%5E2-1%2B0 Now add 0 to both sides to isolate y



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




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-1, 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=1x%5E2%2B2x%2B0 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B2x%2B0%29 Graph of y=1x%5E2%2B2x%2B0. Notice how the vertex is (-1,-1).



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


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



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