+0

# We've got the functions f(x) = ax^2 + 8 and g(x) = 3x^2 - bx

0
187
1

We've got the functions f(x) = ax^2 + 8 and g(x) = 3x^2 - bx

a. For one value of the parameter a is the accompanying graph not an parabola. For which value of a is that the case and what for a function is it then?

b. For the other values of a is the graph of f(x) = ax^2 + 8 a parabola. Which coordinates does the top(vertex?) of each of those graphs have?

c. For which value of a does the graph of f(x) = ax^2 + 8 go through the points (-3, 5)?

d. For which values of a is the graph of f(x) = ax^2 + 8 a parabola opening downward that is more narrow than the ones that belong to g(x) = 3x^2 - bx?

e. The graph of g(x) = 3x^2 - bx cuts the horizontal axis in a point where the x-coordinate is equal to -2. Which value of parameter b belongs to this?

Nov 29, 2018

#1
+2

a) a=0 makes f(x) a constant function

b)  $$\text{the "top" of } f(x) \text{ is }f(0)=0 \\\text{ the "top" of }g(x) \text{ is } f\left(\dfrac b 6\right) =-\dfrac{b^2}{12}$$

c)  $$a(-3)^2 + 8 = 5\\ 9a = -3\\ a=-\dfrac 1 3$$

d)  $$\text{the coefficient of the }x^2 \text{ term controls the narrowness of the parabola}\\ \text{the larger absolute value of the coefficient the narrower the parabola}\\ \text{a negative value produces a downward facing parabola, thus}\\ a < -3 \text{ will produce a downward facing parabola that is narrower than g(x)}$$

e) $$3(-2)^2 - b(-2) = 0\\ 12 +2 b = 0\\ b=-6$$

.
Nov 30, 2018