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.
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:
(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:
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.
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.
That doesn't correspond to what you gave us:
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?
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
where is the vertex and is the directed distance between the vertex and the focus. If is positive, the parabola opens up, and if negative, the parabola opens down.
Since your parabola, has a vertex at and opens up, you can conclude that it is decreasing for and increasing for .
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.