SOLUTION: Given y=3x^2+6x Please explain the process in steps. Thanks! A.) find the y-intercept B.) find the x-intercepts (if any) c.) find the vertex D.) Sketch the parabola.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Given y=3x^2+6x Please explain the process in steps. Thanks! A.) find the y-intercept B.) find the x-intercepts (if any) c.) find the vertex D.) Sketch the parabola.      Log On


   

Question 114528: Given y=3x^2+6x
Please explain the process in steps. Thanks!
A.) find the y-intercept
B.) find the x-intercepts (if any)
c.) find the vertex
D.) Sketch the parabola.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Given:
y=3x%5E2%2B6x+
find:
A.) find the y-intercept…………
find the y-intercept we need to set x equal to 0
y+=3x%5E2%2B6x+
y+=3%2A0%5E2%2B6%2A0+
y+=+0+
B.) find the x-intercepts (if any)
To find the x-intercept, set y equal 0 and solve:
0+=3x%5E2%2B6x+
or

3x%5E2%2B6x+=+0+……………. In this quadratic equation factor x out
x%283x%2B6%29+=+0+…………….find solutions
Above product will be equal to 0 if either one or both factors are equal to 0
So first+solution is x=0
second+solution is 3x%2B6=0, and this will be true if 3x=-6} => x=-6%2F3
Or x+=+-2
Therefore, x_intercepts are at x=0 and x+=+-2

c.) find the vertex

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


y=3+x%5E2%2B6+x%2B0 Start with the given equation



y-0=3+x%5E2%2B6+x Subtract 0 from both sides



y-0=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-0=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-0=3%28%28x%2B1%29%5E2-1%29 Now factor x%5E2%2B2x%2B1 to get %28x%2B1%29%5E2



y-0=3%28x%2B1%29%5E2-3%281%29 Distribute



y-0=3%28x%2B1%29%5E2-3 Multiply



y=3%28x%2B1%29%5E2-3%2B0 Now add 0 to both sides to isolate y



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




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=3, h=-1, 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=3x%5E2%2B6x%2B0 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C3x%5E2%2B6x%2B0%29 Graph of y=3x%5E2%2B6x%2B0. Notice how the vertex is (-1,-3).



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


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



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






D.) Sketch the parabola.
See c.%29