SOLUTION: consider the following quadratic function. Reduce all fractions to the lowest terms. Find the vertex. h(x)=(x-2)(x+6)+16

Algebra ->  Functions -> SOLUTION: consider the following quadratic function. Reduce all fractions to the lowest terms. Find the vertex. h(x)=(x-2)(x+6)+16      Log On


   

Question 105674: consider the following quadratic function. Reduce all fractions to the lowest terms. Find the vertex. h(x)=(x-2)(x+6)+16
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
the vertex. h%28x%29=%28x-2%29%28x%2B6%29%2B16
h%28x%29+=+x%5E2+%2B+6x+%96+2x+%96+12+%2B+16
h%28x%29+=+x%5E2+%2B+4x+%2B+4
h%28x%29+=+%28x+%2B+2%29%5E2+%2B+0+ this is the vertex form, so the vertex is at -2,0

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


y=1+x%5E2%2B4+x%2B4 Start with the given equation



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



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



y-4=1%28x%2B2%29%5E2-1%284%29 Distribute



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



y=1%28x%2B2%29%5E2-4%2B4 Now add 4 to both sides to isolate y



y=1%28x%2B2%29%5E2%2B0 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=-2, 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%2B4x%2B4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2%2B4x%2B4%29 Graph of y=1x%5E2%2B4x%2B4. Notice how the vertex is (-2,0).



Notice if we graph the final equation y=1%28x%2B2%29%5E2%2B0 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x%2B2%29%5E2%2B0%29 Graph of y=1%28x%2B2%29%5E2%2B0. Notice how the vertex is also (-2,0).



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