Objective+11+S3

This is how you do the question for objective 11. There are three types of questions for this objective. You can use you calculator and Graphmatica. I prefer to use Graphmatica. The first type of questions is: Find a and q so that the given points lie on the parabola. y=a(x+2)^2+q ; (-2,-5), (-3,-4)

a) a=1, q=-4 b) a=2,q=5 c)a=1,q=-5 d)a=2,q=7

So all you need to do is use the trial and error. You type each graph in graphmatica and see which graph has both points on it. (-2,-5), (-3,-4). So I typed answer a) and b) and c) and d). Answer c) the image above, lies points (-2,-5),(-3,-4). So its the right question.

Type 2: Write am eqtion that defines the parabola: congruent to y=0.2x^2 :opens up: vertex (5,-1). So to figure this question out you can use the calculator and Grphmatica. I'm going to show you how to nuse Grphmatica agian. a) y= -0.2(x+1)^2-5 b) y= 0.2(x+1)^2-5 c) y= -0.2(x-5) ^2-1 d ) y= 0.2(x-5) ^2-1 You need to find a parabola that is congruent to y=0.2x^2 and opens up and has a vertex of (5,-1) So you type in y= 0.2x^2 in Grphmatica and then you type in the answers above. So this images shows a parabola is congruent and has a vertex of (5,-1). So the answer is b).

Type 3: Write a quadratic function that has the given vertex and passes through the given point.Vertex (-4,-5); Point (-6,-13) a) f(x)= (x-6)^2-2 b) f(x)= 3(x-4)^2+5 c) f(x)= 2(x+3)^2-29 d) f(x)= -3(x-6)^2+13 If you don't know what f(x) is its like y=. You can use your calculator and graphmatica and I used graphmatica to solve this problem. you just type each quesyion in graphmatica and see which one fits the points given to you.

So this image shows which one is right and the points are on the parabola.



That is how you do the questions for Objective 11.