Find A Quadratic Function

[mathdad]

A quadratic function of the form f(x) = ax^2 + bx + c with b^2 - 4ac > 0 may also be written in the form f(x) = a(x - r_1)(x - r_2), where r_1 and r_2 are the x-intercepts of the graph of the quadratic function.

Find a quadratic function whose x-intercepts are -5 and 3 with a = 1; a = 2; a = 5.

1. Why are 3 values of "a" given?
2. I understand the given x-intercepts can be written as (-5, 0) and (3, 0). Is this right?
3. What additional hints can you provide leading to the quadratic function?

 

[Dr.Peterson]

I think it asks three questions with different values of a just to emphasize that a is independent of the intercepts -- it determines the vertical "size" of the graph, as the intercepts determine its horizontal features. So if a=1, you get y = 1(x+5)(x-3); if a=3 you get y = 3(x+5)(x-3); and so on. All these functions have the same intercepts.

Yes, when they say "the x-intercepts are -5 and 3", they could also have said, "the x-intercepts are (-5,0) and (3,0)". These are just different ways to say thing. Some books emphasize the coordinate pairs to make students focus on the graph and how to plot it; others emphasize the numbers (which are sufficient to identify the point, given that it is an x-intercept) to focus on the function itself, and perhaps to avoid redundancy.

Beyond that, all there is to do here is to expand the function and put it in the stated form.

 

[Otis]

A quadratic function of the form f(x) = ax^2 + bx + c with b^2 - 4ac > 0 may also be written in the form f(x) = a(x - r_1)(x - r_2)

... 1. Why are 3 values of "a" given?

Hi mathdad. It's three exercises in one ('a' is a parameter).

2. I understand the given x-intercepts can be written as (-5, 0) and (3, 0). Is this right?

Yes, all x-intercepts are points with a y-coordinate of zero.

3. What additional hints can you provide leading to the quadratic function?

Substitution!

Let r_1 = -5
Let r_2 = 3
Let a = 1

(Swapping the values of r_1 and r_2 makes no difference.)

Write f(x) = a(x - r_1)(x - r_2)

Repeat for the other values of a.

?

 

[mathdad]

I think it asks three questions with different values of a just to emphasize that a is independent of the intercepts -- it determines the vertical "size" of the graph, as the intercepts determine its horizontal features. So if a=1, you get y = 1(x+5)(x-3); if a=3 you get y = 3(x+5)(x-3); and so on. All these functions have the same intercepts.

Yes, when they say "the x-intercepts are -5 and 3", they could also have said, "the x-intercepts are (-5,0) and (3,0)". These are just different ways to say thing. Some books emphasize the coordinate pairs to make students focus on the graph and how to plot it; others emphasize the numbers (which are sufficient to identify the point, given that it is an x-intercept) to focus on the function itself, and perhaps to avoid redundancy.

Beyond that, all there is to do here is to expand the function and put it in the stated form.

Thank you for your informative reply.

 

[mathdad]

Hi mathdad. It's three exercises in one ('a' is a parameter).

Yes, all x-intercepts are points with a y-coordinate of zero.


Substitution!

Let r_1 = -5
Let r_2 = 3
Let a = 1

(Swapping the values of r_1 and r_2 makes no difference.)

Write f(x) = a(x - r_1)(x - r_2)

Repeat for the other values of a.

?

Thank you for separating your reply by the given three parts.

 

[mathdad]

Hi mathdad. It's three exercises in one ('a' is a parameter).

Yes, all x-intercepts are points with a y-coordinate of zero.


Substitution!

Let r_1 = -5
Let r_2 = 3
Let a = 1

(Swapping the values of r_1 and r_2 makes no difference.)

Write f(x) = a(x - r_1)(x - r_2)

Repeat for the other values of a.

?

f(x) = (x + 5)(x - 3)

f(x) = x^2 -3x + 5x - 15

f(x) = x^2 + 2x - 15

Found the first function when a = 1.

I will post the other two functions on separates posts.

 

[mathdad]

The second function is found below.

Let a = 2

The values of r_1 and r_2 remain the same.

f(x) = 2(x + 5) (x - 3)

f(x) = 2(x^2 + 2x - 15)

f(x) = 2x^2 + 4x - 30

 

[mathdad]

Here is the third function when a = 5.

f(x) = 5(x + 5) (x - 3)

f(x) = 5(x^2 + 2x - 15)

f(x) = 5x^2 + 10x - 75

I find this to be an interesting problem. Thanks for your guidance through this question that at first glimpse seems difficult.