Question

Could someone help me solve this one?
What are the coordinates of the vertex for f(x) = x2 + 6x + 13?

a. (4,4)
b. (-4,4)
c. (3,4)
d. (-3,4)

Answer 1

do I just plug in the y coordinate for f(X) and then the x coordinate for x and trial and error?

Answer 2

c

Answer 3

First find your axis of symmetry which is -b/2a . Plugging in numbers does not aid in knowing how to solve this

Answer 4

It is not c.

Answer 5

oh i didn't know that equation thank you what is the b and what is the a?

Answer 6

ax^2 + bx+ c So in the equation a =1 and b = 6. Do you see this?

Answer 7

so the x^2 has a 1 before it and the x has a 6 before it and these are my numbers?

Answer 8

Yes so waht is your axis of symmetry?

Answer 9

so -6/1 = -6?

Answer 10

oops -3

Answer 11

Good job so we know x = -3. Now plug that into the equation to find y...Great Job.

Answer 12

so D is my answer

Answer 13

(-3, 4)?

Answer 14

The trial and error method can work for you if you're on a test and forget the formula though. If the coefficient of x^2 term is negative, then the f(x) that's most positive has to be at the vertex. If it's positive, then the f(x) that's most negative has to be the vertex.

Of course, you have to check that the point given does indeed lie on the graph, so plug in the x-value. This could also just be used to make sure you got the right answer as well.

There are a couple alternate ways. The vertex point is the point at which if you move left or right from it, the height of your function is exactly equal to the other side, so you could equate: f(a-x)=f(a+x) and solve for the value a, which is the x-value of the vertex this way if you're like extreme.

Answer 15

Bingo

Answer 16

Ok, I see what you're saying Kainui. That helps because I hardly ever remember the formula.

Answer 17

\[f(x) = ax^2 + bx + c\]\[f(x) = a\left(x^2 + \dfrac{b}{a}x + \dfrac{c}{a}\right)\]\[f(x) = a\left(x^2 + \dfrac{b}{a}x + \dfrac{c}{a} + \dfrac{b^2}{4a} - \dfrac{b^2}{4a}\right)\]\[f(x) = a\left( \left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2 - 4ac}{4a^2}\right)\]\[f(x) ~\text{is thus maximized or minimized at }\dfrac{-b}{2a} ~\text{by looking at the perfect square}.\]\[\text{The value that the function will then suppose would be }-\dfrac{b^2 - 4ac}{4a}.\]

Answer 18

In fact, for this problem since all the y-values are exactly the same and the vertex is a unique point, we can just plug that in and solve for x.

This question has the same answer as, "Which of the following points lies on the parabola?"

4=x^2+6x+13

Then solve for x that way as well.

Answer 19

In short,\[-\dfrac{-b^2 - 4ac}{4a}\]is written as\[\dfrac{-D}{4a} ~ ~ {\rm or } ~ ~ \dfrac{-\Delta}{4a} ~~~ \rm \text{where D and delta represent the discriminant.}\]
If you have to find out the max/min point of a quadratic, it would be\[\left(-\dfrac{b}{2a}, -\dfrac{D}{4a}\right)\]

Answer 20

Thank you ParthKohli and Kainui. You both explained it perfectly to me! Thanks for helping me pass Algebra 2!If this question comes up again I'll be prepared! Many thanks!

Answer 21

This is one concept that is actually fairly pointless to learn in algebra in my opinion as there is a MUCH better way of doing it later which is much more general when you get to calculus. You simply take the derivative and set it equal to zero since the slope at the vertex is 0. Of course if this makes no sense to you that's alright.

It's still interesting to know how to do this to some degree without calculus though, so I don't mean to sound rude because PK did a great job and I'm sort of just telling you the sneaky way. But as long as you're using your brain to reason it out I'm ok with it.

Answer 22

I just copied this and emailed it to myself for future reference.

Answer 23

... I don't think OP has learned calculus.

Answer 24

I am so "pre" all of this because i am just a sophomore in h.s.

Answer 25

thanks again for the help!

Answer 26

@ParthKohli Yeah I know they haven't, but I think teaching people how to find the vertex of a parabola before calculus is kind of silly. It's like handing someone a pair of scissors and telling them to go mow the lawn.

Answer 27

lol.

That's very subjective though. For example, I love it when algebra is self-sufficient for solving problems. There may be nothing wrong with teaching what the vertex of a parabola is.

The fact that you're able to learn about maximizing/minimizing functions just through algebra is astonishing.

Answer 28

i hate math