Question

MEDAL AND FAN!
Second-degree polynomial is given by f : f(x)=0.5x^2-4x+6
Determine the peak for the corresponding parabola and any intersections with the x-axis.
Determine then value the amount of f (Remember to explain how to get to the answer).

Answer 1

Peak as in highest point? or lowest point?

Answer 2

Find the vertex of the parabola.

Do you know how to do that?

Answer 3

@OrthodoxMan peak meaning top point, which should be the highest point ...

Answer 4

@whpalmer4 no, can you help?

Answer 5

The formula for this parabola is \[y = f(x) = 0.5x^2-4x+6\]which corresponds to the form\[y = ax^2+bx+c\]If \(a>0\) the parabola opens upward, and the vertex is at the bottom.

Are you sure that you have the formula correct? Your question somewhat suggests that the vertex is at the highest point, but with this equation, it will be at the lowest point.

Answer 6

to find the vertex, use the following steps
1. Group x terms in a bracket
2. Factor out coefficient of x^2
3. Divide the other x term by 2 then square it. Take that number, add it and subtract it in the bracket.
4. The subtracted number is to be multiplied by 0.5 and subtracted out of the bracket with the number outside.
5. Factor whatever you have left.
Ex. 0.5(x+____)^2 +/- ____

Answer 7

In any case, if the equation is in the form \[y = ax^2+bx+c\] the x-coordinate of the vertex will be at \[x=-\frac{b}{2a}\]and the y coordinate can be found by plugging that value of \(x\) into \(f(x)\)

Answer 8

@orthodoxman's approach works as well, but is considerably more work :-)

Answer 9

True true :P

Answer 10

Now, if you happen to already have the roots (because it is written in factored form, for example), then it makes getting the vertex quite easy, as the vertex is exactly halfway between the roots.

Answer 11

@whpalmer4, well instead of 0.5 it's says 1/2. But that's about it...

when I use the x=-b/2a, I get 4... What do I do now?

Answer 12

@whpalmer4
Do i do f(4)=0.5(4)^2-4(4)+6?

Answer 13

1/2 = 0.5, of course

Answer 14

Yes, that's the value of y at the vertex

Answer 15

So f(4) = -2 ? @whpalmer4

Answer 16

It does when I compute it...

Answer 17

@whpalmer4, what do i do now? :P

Answer 18

Now you need to find the intersections with the x-axis. Those are the values of x such that f(x) = 0.

Answer 19

@whpalmer4, is the "top point" then (4,-2) ?

Answer 20

well, the vertex is at (4,-2), but it isn't the top point, it's the bottom point. That's why I asked about whether your equation was possibly incorrect. With a negative sign in front of the leading term, we would have a parabola that opens downward, and that would have something that could reasonably be described as a "top point"

Answer 21

But that's all it says. I'll just write that and then see if i get it right. But that would be the point, right? @whpalmer4

Answer 22

You do still have to find the points where the parabola crosses the x-axis...

Answer 23

@whpalmer4 yeah, the intersections. How do I do that? Can you help?

Answer 24

@orthodoxman actually showed you a procedure for doing so. You can also use the formula for the solutions of a quadratic:
if \[y =ax^2+bx+c,\, a\ne 0\]then the solutions are \[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 25

@whpalmer4 , I know y should 0. Because it's the intersections that's on the x-axis we need to find... right?

Answer 26

yes, y =0 at those points.

Answer 27

@whpalmer4 Thanks! I'll do that first!

Answer 28

@whpalmer4
I got x=1 or x=1/3 .... do you know if that's right?

Answer 29

plug in the values in the original equation. do you get y = 0 if so?

\[y = \frac{1}{2}x^2-4x+6\]
\[y = \frac{1}{2}(1)^2-4(1)+6\]\[y=\frac{1}{2}+2\]Nope, not correct.

Answer 30

Also, the two solutions are symmetrical about the vertex. If the vertex is at x = 4, one solution is at x-a, the other at x + a, for some value of a.

Answer 31

Find 2 x-intercepts:
y = x^2/2 - 4x + 6 = 0 = x^2 - 8x + 12 = 0.
Both roots are positive. Compose factor pairs of c = 12. Proceeding: (1, 12)(2, 6). This last sum is 2 + 6 = 8 = -b. Then, the 2 real root are 2 and 6.
Min at x = -b/2a = 4/1 = 4
Ymin. = 8 - 16 + 6 = -2.

Answer 32

@whpalmer4 i have no idea what i'm doing now

Answer 33

@thu1935 yeah! I've done that one :-)

Answer 34

Why don't you show me the work you did with the quadratic formula that gave you the answers of 1 or 1/3? Let's find out where you made the mistake

Answer 35

First, what are you using for a, b, c?

Answer 36

I'm using a= 1/2, b=-4 and c=6

Answer 37

@whpalmer4

Answer 38

@whpalmer4, you don't know what I did wrong ?

Answer 39

the denominator is \(2a\) not \(2ac\)

Answer 40

@whpalmer4 oh, yeah. Crap.

Answer 41

So it's 6 and 2?

Answer 42

Yup! it's right! @whpalmer4

Answer 43

And 6 and 2 are symmetrically located about 4.

Graph looks like this:

Answer 44

Anyway. Thanks a bunch, @whpalmer4 and @OrthodoxMan!

Answer 45

You're welcome!