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.Com
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)   (Show Source): You can put this solution on YOUR website!
This is not the full solution.






-----------you can read the vertex and solve for the roots.




--------------you can read the vertex and solve for the roots.
Answer by ewatrrr(24785)   (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)   (Show Source): You can put this solution on YOUR website!

given:
AND

first solving for the x intercepts using the quadratic formula:





solutions:


->if the solutions are complex, the quadratic does not have x-intercepts
see it on a graph:











solutions:
=>
or
=>
->if the solutions are complex, the quadratic does not have x-intercepts

see it on a graph:



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



=>if =>
=>if =>

The average of those is the coordinate of the vertex.


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




so we have the vertex at (,)



do same with other equation


solutions:
if =>
if =>

average is





so we have the vertex at (,)





Answer by MathTherapy(10551)   (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, -----------you can read the vertex and solve for the roots"
IS NOT the vertex form of the equation of: 


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.


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!!

                                          

Removing the constant, the above becomes: 

                                          
                                           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). 
      
                                          

Removing the constant, the above becomes: 

                                          
                                           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).
RELATED QUESTIONS

Explain how you would find the x-intercepts and y-intercept for quadratic... (answered by stanbon)
Please help!!! I have tried solving the following problems, but can't seem to... (answered by checkley75)
1. Find two positive real numbers whose sum is 40 and whose product is a... (answered by JimboP1977)
please assist me in solving this problem by using the quadratic formula... (answered by Edwin McCravy)
Solve the following equation for x by using the quadratic formula: 2x^2 + 8x + 3 =... (answered by stanbon)
Solve the following equation for X by using the quadratic formula. 4x^2+2x-4=0 (answered by solver91311)
Solve the following equation for x by using the quadratic formula... (answered by Edwin McCravy)
Sketch the graph of the following functions by using the conditions of concavity... (answered by MathLover1)
Use the quadratic formula to determine the x-intercepts (if any) of the following... (answered by scott8148)