MHB How to Tackle Pre-Calculus Problems Effectively?

[Samanthalovesmath]

How would I approach this?

 

[DaalChawal]

Read quadratic chapter once these are direct questions from theory.

 

[HOI]

You are told that "A quadratic equation of the form $f(x)= ax^2+ bx+ c$ with $b^2- 4ac$ can also be written $f(x)= a(x- r_1)(x- r_2)$ where $r_1$ and $r_2$ are the x-intercepts of the graph of the quadratic function".

Then you are asked to "(a) Find a quadratic function whose x-intercepts are -5 and 1 with a= 1, a= 2, a= -2, and a= 6."

Do you understand that these are 4 different problems with four different answers? Since you are told that the "x-intercepts are -5 and 1" you know immediately that the function is f(x)= a(x- (-5))(x- 1)= a(x+ 5)(x- 1). Then just put in the four different values of a.

(b) How does the value of a affect the intercepts?
Since you have just written four different functions with different values of a but exactly the same x-intercepts, it should be clear that the value of a does NOT affect the x-intercepts. However, this question might be referring to the y-intercept, where x= 0, as well. f(0)= a(0+ 5)(0- 1)= -5a. How does the value of a affect that?

(c) How does the value of a affect the axis of symmetry?
Do you know what the "axis of symmetry" is? You should know that the graph of a quadratic function is a parabola. By "completing the square" you can always write the quadratic as $f(x)= a(x- b)^2+ c$ for some numbers a, b, and c. Then the "axis of symmetry" is x= b because $y= (x- b)^2$ is symmetric about x= b. Here the function is $f(x)= a(x+5)(x- 1)= a(x^2+5x- x- 5)= a(x^2+ 4x- 5)$. Do you know how to "complete the square"? How does the value of a affect that?

d) How does the value of a affect the vertex.
Once you have "completed the square" so that the function is written $f(x)= a(x- b)^2+ c$, the "vertex" is at (b, c).

e) Compare the x-coordinate of the vertex with the midpoint of the x-intercepts. What might you can conclude?
You were told that the x-intercepts were -5 and 1. The midpoint of that interval is (-5+ 1)/2= -4/2= -2. What was the x-coordinate of the vertex?

 

[karush]

I started a collection of Pre Calculus problems with MHB replies
I am sure it will help a lot as the list grows longer
https://dl.orangedox.com/QS7cBvdKw55RQUbliE

Mahalo