SOLUTION: Sketch the following functions by first solving for the x intercepts using the quadratic formula.  If there are no x intercepts then use the method explained in class...this makes

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Sketch the following functions by first solving for the x intercepts using the quadratic formula.  If there are no x intercepts then use the method explained in class...this makes      Log On


   

Question 1196808: Sketch the following functions by first solving for the x intercepts using the quadratic formula.  If there are no x intercepts then use the method explained in class...this makes use of the fact that raising or lowering a parabola by some amount does not change the x coordinate of the vertex.  So...remove the constant from the right side, factor what's left to determine x intercepts.  The average of those is the x coordinate of the vertex.  Then substitute that back into the original equation to solve for the y coordinate of the vertex. 
y = 3x^2+15x+30   AND   y = -x^2+4x-8
Found 4 solutions by josgarithmetic, ewatrrr, MathLover1, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
This is not the full solution.

3%28x%5E2%2B5x%2B10%2B%285%2F2%29%5E2-%285%2F2%29%5E2%29
3%28x%5E2%2B5x%2B%285%2F2%29%5E2%2B10-%285%2F2%29%5E2%29
3%28%28x%2B5%2F2%29%5E2%2B40%2F4-25%2F4%29
3%28%28x%2B5%2F2%29%5E2%2B15%2F4%29
3%28x%2B5%2F2%29%5E2%2B45%2F4
y=3%28x%2B5%2F2%29%5E2%2B11%261%2F4-----------you can read the vertex and solve for the roots.

-%28x%5E2-4x%2B8%29
-%28x%5E2-4x%2B4-4%2B8%29
-%28%28x-2%29%5E2%2B4%29
y=-%28x-2%29%5E2-4--------------you can read the vertex and solve for the roots.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
Finding the Vertex Form of the EQ by completing the Square:
y = 3x^2+15x+30  = 3(x^2 + 5x + 10) = 3((x+ 5/2)^2 -25/4 + 10) = 3(x+5/2)^2 + 45/4
y = 3(x+5/2)^2 + 45/4  has V(-5/2, 45/4)
Wish You the Best in your Studies.


Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

given:
y+=+3x%5E2%2B15x%2B30 AND y+=+-x%5E2%2B4x-8

first solving for the x intercepts using the quadratic formula:
y+=+3x%5E2%2B15x%2B30
x=%28-15%2B-sqrt%2815%5E2-4%2A3%2A30%29%29%2F%282%2A3%29+
x=%28-15%2B-sqrt%28225-360%29%29%2F6+
x=%28-15%2B-sqrt%28-135%29%29%2F6+
x=%28-15%2B-i%2Asqrt%28135%29%29%2F6+
solutions:
x+=+-5%2F2+%2B+%28i%2Asqrt%2815%29%29%2F2
x+=+-5%2F2+-+%28i%2Asqrt%2815%29%29%2F2
->if the solutions are complex, the quadratic does not have x-intercepts
see it on a graph:
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+40%2C3x%5E2%2B15x%2B30%29+


y+=+-x%5E2%2B4x-8

x=%28-4%2B-sqrt%284%5E2-4%2A%28-1%29%2A%28-8%29%29%29%2F%282%2A%28-1%29%29+
x=%28-4%2B-sqrt%2816-32%29%29%2F%28-2%29+
x=%28-4%2B-sqrt%28-16%29%29%2F%28-2%29+
x=%28-4%2B-4i%29%2F%28-2%29+
x=%28-2%2B-2i%29%2F%28-1%29+
x=-%28-2%2B-2i%29+
solutions:
x=-%28-2%2B2i%29+ =>x=2-2i+
or
x=-%28-2-2i%29+ =>x=2%2B2i+
->if the solutions are complex, the quadratic does not have x-intercepts

see it on a graph:
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+-x%5E2%2B4x-8%29+


using factoring method:
remove the constant from the right side, factor what's left to determine+x intercepts

0=+3x%5E2%2B15x
0+=+3x%28x+%2B+5%29
=>if 3x=0=>x=0
=>if x%2B5=0=>x=-5

The average of those is the x coordinate of the vertex.
%280%2B%28-5%29%29%2F2=-5%2F2=-2.5

substitute that back into the original equation to solve for the y coordinate of the vertex

y=+3%2A%28-2.5%29%5E2%2B15%2A%28-2.5%29
y=+-18.75

so we have the vertex at (-2.5,-18.75)
+graph%28+600%2C+600%2C+-10%2C+10%2C+-20%2C+40%2C3x%5E2%2B15x%29+


do same with other equation
0+=+-x%5E2%2B4x
0+=+-x%28x-4%29
solutions:
if 0+=+-x =>x=0
if 0+=+x-4=>x=4

average is %280%2B4%29%2F2=2

y=+-2%5E2%2B4%2A2
y=+-4%2B8
y=+4

so we have the vertex at (2,4)

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+40%2C-x%5E2%2B4x%29+



Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
Sketch the following functions by first solving for the x intercepts using the quadratic formula.  If there are no x intercepts then use the method explained in class...this makes use of the fact that raising or lowering a parabola by some amount does not change the x coordinate of the vertex.  So...remove the constant from the right side, factor what's left to determine x intercepts.  The average of those is the x coordinate of the vertex.  Then substitute that back into the original equation to solve for the y coordinate of the vertex. 
y = 3x^2+15x+30   AND   y = -x^2+4x-8
First, as stated by one of the people who responded, y=3%28x%2B25%2F4%29%5E2-75%2F4-----------you can read the vertex and solve for the roots"
IS NOT the vertex form of the equation of: matrix%281%2C3%2C+3x%5E2+%2B+15x+%2B+30%2C+%22=%22%2C+0%29

matrix%281%2C3%2C+y%2C+%22=%22%2C+3x%5E2+%2B+15x+%2B+30%29
Although prompted to use the quadratic equation formula to find the x intercepts, it's not necessary. Calculating the discriminant, , it's absolutely clear that this equation's solutions are IMAGINARY, and so, DOES NOT
have any x-intercepts.

matrix%281%2C3%2C+y%2C+%22=%22%2C+-+x%5E2+%2B+4x+-+8%29
Although prompted to use the quadratic equation formula to find the x intercepts, it's not necesary. Calculating the discriminant, , it's absolutely clear that this equation's solutions are IMAGINARY, and so, DOES NOT
have any x-intercepts.

I have NO IDEA what method was explained in YOUR class!!

                                          matrix%281%2C3%2C+y%2C+%22=%22%2C+3x%5E2+%2B+15x+%2B+30%29

Removing the constant, the above becomes: 

                                          matrix%282%2C3%2C+0%2C+%22=%22%2C+3x%5E2+%2B+15x%2C+0%2C+%22=%22%2C+3x%28x+%2B+5%29%29
                                           0 = 3x     or      0 = x + 5
                                           0 = x      or    - 5 = x 
                               x-intercepts: 0 and - 5, or (0, 0) and (- 5, 0)
x-coordinate of vertex of:  
y-coordinate of vertex of: 
Coordinates of vertex of: 3x2 + 15x + 30: (- 2.5, 11.25). 
      
                                          matrix%281%2C3%2C+y%2C+%22=%22%2C++-+x%5E2+%2B+4x+-+8%29

Removing the constant, the above becomes: 

                                          matrix%282%2C3%2C+0%2C+%22=%22%2C+-+x%5E2+%2B+4x%2C+0%2C+%22=%22%2C+-+x%28x+-+4%29%29
                                           0 = - x     or      0 = x - 4
                                           0 = x       or      4 = x 
                               x-intercepts: 0 and 4, or (0, 0) and (4, 0)
x-coordinate of vertex of:  
y-coordinate of vertex of: 
Coordinates of vertex of: - x2 + 4x - 8: (2, - 4).