Help with vertex problem

mtcmath

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Joined
Nov 22, 2009
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5
Can you please help me to work this problem: solve and graph:

5x^2 - 10x + 26
 
mtcmath said:
Can you please help me to work this problem: solve and graph:

5x^2 - 10x + 26

Write the equation in the following form

y = A * (x - h)[sup:39puks5h]2[/sup:39puks5h] + k where A = constant and the co-ordinate of the vertex is (h,k)

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
v: (--10/2(5), f(x)= 1)

5(1)^2 - 10(1) + 26
5 - 10 + 26 = 21

5x^2 - 10x + 26 = x = --10+/- (square root: (-10)^2-4(5)(26)
_________________________________
2(5)

x= 10 +/- 2 (square root: 105i
_______________________
10

= 1 +/- 1 (square root: 105i
_________________________
5
 
Hello, mtcmath!

What do you mean by "solve"?


Solve and graph: .\(\displaystyle y \:=\:5x^2 - 10x + 26\)

If you mean "Find the vertex", there's formula for it.

\(\displaystyle \text{Given the parabola: }\:y \:=\:ax^2 + bx + c\)

. . \(\displaystyle \text{the vertex is at: }\:x \;=\;\frac{\text{-}b}{2a}\)


\(\displaystyle \text{We have: }\:a = 5,\;b = \text{-}10,\;c = 26\)

\(\displaystyle \text{Then we have: }\:x \:=\:\frac{\text{-}(\text{-}10)}{2(5)} \:=\:1\)

. . \(\displaystyle \text{and: }\;y \:=\:5(1^2) - 10(1) + 26 \:=\:21\)

\(\displaystyle \text{The vertex is at: }\:(1,21)\)


\(\displaystyle \text{Find a few more points and graph the parabola.}\)

Code:
        |
  *     |                 *
   *    |                *
     *  |              *
  (0,26)o           o(2,26)
        |     o
        |   (1,21)
        |
        |
- - - - + - - + - - - - - - - - -
        |     1
 
I'm sorry for not explaining myself completely: I did get the vertex of ( 1,21) and y-intercept: (0, 26), but I also had to find the x-intercepts: that is why I did the quadratic function to solve for x, but I don't think I did it right.
 
v: (--10/2(5), f(x)= 1)

5(1)^2 - 10(1) + 26
5 - 10 + 26 = 21
Code:
5x^2 - 10x + 26 = x = --10+/- (square root: (-10)^2-4(5)(26)
                                   _________________________________
                                                          2(5)

x= 10 +/- 2 i (square root: 105)
     _______________________
                        10

=  1 +/- i [(square root: 105)]/5
[/quote]
If you are looking for x-intercept - your calculations are correct (except as indicated above)

This function (graph) does not cross x-axis - as evidenced by i in the expressions for roots.
 
thank you, and one last question please, when you have a negative square root, how do you know to have the i on the outside or inside the square root, such as: the square root of ?-28, would it be written as 2?7i or 2i?7
 
mtcmath said:
thank you, and one last question please, when you have a negative square root, how do you know to have the i on the outside or inside the square root, such as: the square root of ?-28, would it be written as 2?7i or 2i?7
Remember:

\(\displaystyle i^2 = -1\)

So:

\(\displaystyle \sqrt{-28} = \sqrt{(-1)\cdot 7 \cdot 4} = \sqrt{(i^2)\cdot 7 \cdot 2^2} = i2\sqrt{7}\)
 
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