Functions Pre-Calculus Class

[NerdyAndrea]

Hello everyone:

I am in my first math class in 10 years. I am struggling but keeping my head above water. I am having an awful time with functions. I am supposed to sketch the graph of a quadriatic function without using a graphin utility sadly that leaves out my TInspire. I have to identify the vertex, axis of symmetry, and x-intercepts.

I have been reading the instructions in my math book and googled functions, and such, but I am not sure how to apply what I'm reading. I guess I must need step by step explanations of why things are the way they are.

I understand that the function of x is y, I am just kind of baffled after that. Here is the problem:

h(x)=x^2 - 8x + 16

If anyone can explain this to me, I'll be incredibly greatful! Thank you for taking the time to look at my problem in advance. I am a pre-med student and am wanting to get good grades, and trying very hard or that!

Thanks!

NerdyAndrea

 

[mmm4444bot]

I am not sure how to apply what I'm reading

I must need step by step explanations

I am just kind of baffled

Hi Andrea:

I could certainly type up a bunch of stuff, but would you understand it?

If not, then we'd both be wasting our time.

Rather than asking us to guess why you do not understand something, I suggest that you post specific questions.

Tell us what it is specifically that you do not understand about the definition of "vertex".

Tell us what it is specifically that you do not understand about the definition of "axis of symmetry".

Likewise, what is it specifically that you don't understand about x-intercepts?

In other words, if you read a sentence somewhere that baffles you, then ask somebody about that sentence.

h(x) = x^2 - 8x + 16

We see that function h is defined by a quadratic polynomial.

The x-intercepts become obvious, if we (1) factor the polynomial and (2) set each factor equal to zero and solve for x.

So, my question for you is: "Do you know how to factor x^2 - 8x + 16 ?"

(Factoring "easy" polynomials, like this one, is a skill taught in beginning algebra. In other words, this skill is prerequisite knowledge, for a pre-calculus course.)

There is a formula to calculate the x-coordinate of the vertex point. This formula is stated in terms of the coefficients A and B.

Given y = Ax^2 + Bx + C, the x-coordinate of the vertex is -B/(2A).

Of course, once you have a number for x, finding the corresponding y-coordinate is simply a matter of evaluating h(x).

So, my question for you is: "Do you know what the values are for the three coefficients A, B, and C, in the given polynomial that defines h(x) ?"

The axis of symmetry for any polynomial is simply a line that "divides" the parabola into two symmetrical halves. For parabolas that open up or down (versus other types which open left or right or rotated), the axis of symmetry is always a vertical line that passes through the x-coordinate of the vertex point.

So, my question for you is: "Do you know how to write the equation for a vertical line ?"

?

I am in my first math class in 10 years. To me, this statement raises a red flag.

Are you telling us that you have not done any mathematics for 10 years, and that you have now somehow been allowed to enroll in a precalculus course?

Is this one of those on-line courses? For your sake, I certainly hope not.

Without knowing what you already know, I can't really tutor you in the absence of specific questions. However, I'd be willing to post the step-by-step instructions for a similar exercise (without explanations), if you think that would help.

I cannot, in this forum, teach you several weeks worth of material that is covered in beginning and intermediate algebra (i.e., the prerequisite knowledge needed before studying precalculus).

 

[NerdyAndrea]

Hi there mmm4bot:

I know how to factor, and I have the problem factored, I know the vertex is (4,0) and that factored out I have (x - 4)(x - 4) need to know how I got the vertex,a nd I understand my parabola is postive. No it's not an online course, I atually tested into pre-calculus and I have no idea of how that happened, other than that I probably remembered a lot, and now I judt don't understand how I got there. I am fine on other things, this is really my only shaky ground. I am doing fine with polynomials adding, subtracting and dividing and imaginary numbers. It was too genreal what I posted earlier, and no, I don't need weeks of things, I'm jut shaky on some. And yes I truly haven't had a math class in over ten years, but I remember a lot.

Thanks for asking for clarification, I'll narrow it dow more the next time.

A

 

[mmm4444bot]

I know the vertex is (4,0)

need to know how I got the vertex

Hey A,

Okay, I interpret the above statements to mean that somebody gave you the vertex coordinates (4, 0), and you're asking how to find them.

We can use this formula, to first determine the x-coordinate at the vertex point of any parabola that opens up or down.

x-coordinate = -B/(2A)

(I remember this formula by realizing that it's what remains in the Quadratic Formula when the discriminant is zero.)

In the polynomial that defines function h, we see that A = 1 and B = -8.

x-coordinate = -(-8)/[(2)(1)]

This simplifies to 8/2, which is 4, so the x-coordinate is 4.

To find the corresponding y-coordinate, we simply evaluate h(4) because y = h(x), which literally means that the variable h(x) is the y-coordinate for each x-coordinate along the Real number line (x-axis).

h(4) = 0

Vertex: (4, 0)

The axis of symmetry is the vertical line that passes through the vertex.

Do you remember the equation for a vertical line? It's simply x = c, where c is the distance from the line to the y-axis.

As far as x-intercepts go, in this exercise, you know that the parabola opens upward AND that the vertex is on the x-axis. So what does that information tell you about locations where the parabola meets the x-axis?

Were we to be wiithout this information, we would need to solve the equation x^2 - 8x + 16 = 0 for x. Obviously, these solution(s) give the location of x-intercepts. Do you understand why? Think: y = h(x), and all points on the x-axis look like (x,0).

Graphing by hand requires evaluating a few more values of x, to obtain a few more points to plot, other than the vertex.

The y-intercept is always easy to calculate, right? What's h(0) ?

You can pick some arbitrary values for x, to find more points to plot.

Let x be the numbers -1, 1, 2, and 3.

Now calculate the numbers h(-1), h(1), h(2), and h(3), as these are the corresponding y-values.

I realize that I'm repeating myself, but I want you to realize the following.

The equation h(-44) = 2304 means that the point (-44, 2304) is on the parabola.

The equation h(13/17) = 3025/289 means that the point (13/17, 3025/289) is on the parabola.

Abstractly (representing all possibilities), the equation h(x) = y means that a point is (x, h(x)), as we let x take on all Real values. You only need to figure out a few, to sketch a rough graph! And once you realize the above, it becomes clear what you need to do.

Plot these four points, which all lie to the left of the vertex.

Use symmetry to plot four more points to the right of the vertex, and then draw the smooth curve.

As an example of how to use symmetry: h(3) = 1, so, we know that the point (3, 1) is part of the parabola. Since 3 is one unit to the left of the vertex, symmetry tells us that the curve behaves the same one unit to the right of the vertex.

In other words, once we know (3, 1) we also know (5, 1) because of symmetry.

As far as future posts go, yes, the more specific the information you provide, the better the responses that you will get. I came away from your original post speculating that you didn't really know very much at all.

Let us know, if you still have questions about this exercise.

Cheers

 

[NerdyAndrea]

It meets the parabola is int he right quadrant, and the opening is probably of equal spacing since the number is positive. If I remember correctly from my notes and instructions, if the x coordinate was a negative the parabola would have opened wider.

Yep remember that, it's undefined slope when X=C

Okay thank you that explanation of getting there finally made some sense to me. I kept going over my book and notes and it wasn't there. Now I can finish my homework up! Yay!

Andrea