# How to find vertex of a graph or equation

• Jun 6th 2008, 11:31 AM
snakeman11689
How to find vertex of a graph or equation
I think the title say's it all. I have done a problem which requires me to give the vertex, but I'm not quite sure how to find it. From a graph that I interpreted so far I have x-intercepts:1,3 and y-intercept: -3. I don't know how to find the vertex for an equation also. Can someone teach me step by step on how to find the vertex. I'm not a math person, so I'm not that good with math.
• Jun 6th 2008, 11:33 AM
Mathstud28
Quote:

Originally Posted by snakeman11689
I think the title say's it all. I have done a problem which requires me to give the vertex, but I'm not quite sure how to find it. From a graph that I interpreted so far I have x-intercepts:1,3 and y-intercept: -3. I don't know how to find the vertex for an equation also. Can someone teach me step by step on how to find the vertex. I'm not a math person, so I'm not that good with math.

What is the equation?
• Jun 6th 2008, 11:39 AM
o_O
Seeing how we don't have the graph right now, we can only guess. But I think we have a good idea of what your parabola looks like. Since x = 1 and 3 are x-intercepts, you can say that (x-1) and (x-3) are factors of your quadratic equation:
$\displaystyle y = a(x - 1)(x-3)$ (if you plug in 1 or 3, you'll get y = 0 which is what an x-intercept is); a is some constant that is generally associated with this form of a quadratic equation

You also know that (0, -3) (i.e. x = 0 and y = -3) is on your graph:
$\displaystyle -3 = a(0 - 1)(0-3)$

So you can solve for a.

Now, you know the vertex will occur in the midpoint between x = 1 and x = 3 (because quadratic equations are symmetrical) so x = 2 is where your vertex will be. Now plug in x = 2 to your formula to find the vertex and you'll be done.
• Jun 6th 2008, 01:51 PM
snakeman11689
Quote:

Originally Posted by Mathstud28
What is the equation?

f(x)= -(x-3)^2 + 2
• Jun 6th 2008, 01:54 PM
snakeman11689
Do you guys know how to find increasing and decreasing intervals for graphs. Let's just start on something simple.

some function like f(x)= x^2 + 2 <--- the second 2 is not sticked to the exponential.

Since, it forms a parabola by going up the y axis by 2. How do you do increasing and decreasing intervals??? Such as (2,infinity)=increasing, (-infinity, 2)= decreasing.
• Jun 6th 2008, 01:56 PM
o_O
Quote:

Originally Posted by snakeman11689
f(x)= -(x-3)^2 + 2

That doesn't correspond to what you gave us:
$\displaystyle f(3) = -(3-3)^{2} + 2 = 2 \neq 0$ indicating that x = 3 is not an x-intercept.

Quote:

Originally Posted by snakeman11689
Do you guys know how to find increasing and decreasing intervals for graphs. Let's just start on something simple.

some function like f(x)= x^2 + 2 <--- the second 2 is not sticked to the exponential.

Since, it forms a parabola by going up the y axis by 2. How do you do increasing and decreasing intervals??? Such as (2,infinity)=increasing, (-infinity, 2)= decreasing.

The intervals refer to the x values in which the function is increasing or decreasing. I'm assuming you don't know any calculus so you'll have to visually determine this. For your parabola, for what x values is the function decreasing? For what x values is the function increasing?
• Jun 6th 2008, 02:10 PM
snakeman11689
Quote:

Originally Posted by o_O
That doesn't correspond to what you gave us:
$\displaystyle f(3) = -(3-3)^{2} + 2 = 2 \neq 0$ indicating that x = 3 is not an x-intercept.

The intervals refer to the x values in which the function is increasing or decreasing. I'm assuming you don't know any calculus so you'll have to visually determine this. For your parabola, for what x values is the function decreasing? For what x values is the function increasing?

Umm that problem is another problem.

And for interval thingy, would decreasing= [0, -infinity). I'm really not sure what's going on with intervals.
• Jun 6th 2008, 02:12 PM
Reckoner
Quote:

Originally Posted by snakeman11689
Do you guys know how to find increasing and decreasing intervals for graphs. Let's just start on something simple.

some function like f(x)= x^2 + 2 <--- the second 2 is not sticked to the exponential.

Since, it forms a parabola by going up the y axis by 2. How do you do increasing and decreasing intervals??? Such as (2,infinity)=increasing, (-infinity, 2)= decreasing.

If a function is increasing over an interval, its derivative will be positive, and if it is decreasing it will be negative. I assume you know how to do differentiation, since you posted in calculus.

Of course, you can also use the equation of the parabola to determine intervals on which it is increasing or decreasing. In standard form, a parabola can be represented by the equation

$\displaystyle y - k = \frac1{4p}(x - h)^2$

where $\displaystyle (h,\;k)$ is the vertex and $\displaystyle p$ is the directed distance between the vertex and the focus. If $\displaystyle p$ is positive, the parabola opens up, and if negative, the parabola opens down.

Since your parabola, $\displaystyle f(x) = x^2 + 2 = (x - 0)^2 + 2$ has a vertex at $\displaystyle (0,\;2)$ and opens up, you can conclude that it is decreasing for $\displaystyle x\in(-\infty,\;0)$ and increasing for $\displaystyle x\in(0,\;\infty)$.
• Jun 6th 2008, 02:19 PM
snakeman11689
Quote:

Originally Posted by Reckoner
If a function is increasing over an interval, its derivative will be positive, and if it is decreasing it will be negative. I assume you know how to do differentiation, since you posted in calculus.

Of course, you can also use the equation of the parabola to determine intervals on which it is increasing or decreasing. In standard form, a parabola can be represented by the equation

$\displaystyle y - k = \frac1{4p}(x - h)^2$

where $\displaystyle (h,\;k)$ is the vertex and $\displaystyle p$ is the directed distance between the vertex and the focus. If $\displaystyle p$ is positive, the parabola opens up, and if negative, the parabola opens down.

Since your parabola, $\displaystyle f(x) = x^2 + 2 = (x - 0)^2 + 2$ has a vertex at $\displaystyle (0,\;2)$ and opens up, you can conclude that it is decreasing for $\displaystyle x\in(-\infty,\;0)$ and increasing for $\displaystyle x\in(0,\;\infty)$.

oh i kinda get it know, it has to deal with the vertex to find the intervals rite.
• Jun 6th 2008, 06:23 PM
snakeman11689
how do we do the intercepts of a equation

f(x)= -(x-3)^2 + 2<--- The +2 is separate from the exponential.
• Jun 6th 2008, 07:30 PM
Reckoner
Quote:

Originally Posted by snakeman11689
how do we do the intercepts of a equation

f(x)= -(x-3)^2 + 2<--- The +2 is separate from the exponential.

The x-intercepts occur when $\displaystyle y = 0$, so solve the equation $\displaystyle f(x) = 0$ for $\displaystyle x$. The y-intercepts occur when $\displaystyle x = 0$, so simply evaluate $\displaystyle f(0)$. For example:

$\displaystyle \text{Find the }x\text{ and }y\text{ intercepts of the graph of } f(x) = x^2 - 5x + 6$

The y-intercepts will occur at $\displaystyle \left(0,\;f(0)\right) = (0,\;6)$. The x-intercepts will occur when $\displaystyle y = 0$:

$\displaystyle f(x) = x^2 - 5x + 6 = 0$

$\displaystyle \Rightarrow(x - 3)(x - 2) = 0$

$\displaystyle \Rightarrow x = 3 \text{ or } x = 2$

So the x intercepts are at $\displaystyle (3,\;0)\text{ and }(2,\;0)$.
• Jun 6th 2008, 07:34 PM
snakeman11689
hmm, the answer in the back of the book gave me x-int: 3 +- squareroot 2
y-int: -7
• Jun 6th 2008, 07:37 PM
Reckoner
Quote:

Originally Posted by snakeman11689
hmm, the answer in the back of the book gave me x-int: 3 +- squareroot 2
y-int: -7

The problem I gave was just an example I made up, not the one you posted. You should be able to solve yours by the same methods.
• Jun 6th 2008, 07:50 PM
snakeman11689
Quote:

Originally Posted by Reckoner
The problem I gave was just an example I made up, not the one you posted. You should be able to solve yours by the same methods.

hmm, i got your way of getting the y int. But i kinda have a hard time getting the x-intercept

I used a different method for finding the x-int

-(x-3)^2 +2
(x-3)(x-3)+2
-x^2-6x+9+2
-x^2-6x+11
or should it be
-x^2+6x-7

After all of this, should i use the quadratic formula, since the x-int answer turned out to have a square root with it.
• Jun 6th 2008, 07:57 PM
Reckoner
Quote:

Originally Posted by snakeman11689
After all of this, should i use the quadratic formula, since the x-int answer turned out to have a square root with it.

You could use the quadratic formula, but it would be easiest to leave it in the factored form and solve from there:

$\displaystyle -(x-3)^2 + 2 = 0$

$\displaystyle \Rightarrow(x-3)^2 = 2$

$\displaystyle \Rightarrow x-3 = \pm\sqrt2$

$\displaystyle \Rightarrow x = 3\pm\sqrt2$