Help With Transformations Of Quadratic Functions

Quadratic functions and transformations?

do you mean c= 0.000015x ^2 -0.03x+35 ?
to minimize the cost, we need to take the derivative and set equal to zero.

c' = .00003x - .03
0 = .00003x - .03
.03 = .00003x
x = .03 / .00003 = 1000

so our x value is 1000. We just need to plug this into the equation for c to find the cost.
c = .000015 * (1000^2) - .03 (1000) + 35
c = .000015 x 10^6 - .03 x 10^3 + 35
c = 15 - 30 + 35
c = 20

so the cost is 20.

Hope this helps.

Use transformations and the zeros of the quadratic function f(x) = (x+4) (x-2) to determine the zeros of :?

Hi,

The zeros of the function y= 3f(x) are still at x = -4 and x = 2 just like the original f(x). The 3 out front makes every y value three times the previous y value, so when the y value is zero at the x intercepts, then 3 * 0 = 0. Therefore, the x intercepts are the only points on y = 3f(x) that are the same points as what were on y = f(x).

The zeros of the function y= f(2x) are at x = -2 and x = 1. The new function f(2x) = (2x +4)(2x - 2) solves as 2x + 4 = 0 and 2x - 2 = 0. These x intercepts are x = -2 and x = 1.

I hope that helps!! :-)

Using transformations to graph quadratic functions help???

1. Start with f(x)=x^2. Normal parabola opening up.
stretch it by a factor of 2... 2x^2 the vertex is still at (0,0), f(1)=2
translate 3 units to the left 2(x+3)^2, the vertex is now at (-3,0), f(1)=32 (it's no longer 1 unit from the vertex)
g(x)=2(x+3)^2

2. g(x)=x^2-2
this is a parabola that has been shifted down 2 units with no stretch, compression or reflection

the vertex is at (0,-2) and the intercepts are at (-√2,0),(√2,0), f(1)=-1

It's easier to see the transformations if you work in vertex form.
y (or f(x)) = a(x-h)^2+k
a is the stretch or compression factor if 0if a>1 it is a stretch
if a is negative, it has been reflected across the x-axis
(h,k) is the vertex. the - sign is part of the formula so if the vertex is -4,2 it would become (x+4)
H: to shift to the right, it is - the number of units (because subtracting a positive number is still subtracting), to the left + the number of units (subtracting a negative number is the same as adding)
K: add to shift up, subtract to shift down

you don't see too much horizontal compression or stretching since it looks like the vertical ones but if you added a coefficient to x inside the parenthesis you would get a compression if the coefficient is >1 or a stretch if it is less than 1 (but greater than 0)

to reflect it across the y-axis, the x values change from x to -x (anything in QI moves to QII and QII moves to QI. In vertex form that is (x-h) goes to (h-x) or (x+h) goes to (-x-h) (Note do not factor out the -1 in this case)

Using transformations to graph quadratic functions?

Use the description to write each quadratic function in vertex form.
1.) The parent function f(x)=x^2 is vertically stretched by a factor of 2 and translated 3 units left to create g.
Using the graph of f(x)=x^2 as a guide, describe the transformations, and then graph each function.
Select the appropriate transformation(s) for 2-6
a. vertical translation up
b. vertical translation down
c. horizontal translation left
d. horizontal translation right
e. reflection across x-axis
f. reflection across y-axis
g. vertical stretch
h. vertical compression
i. horizontal stretch
j. horizontal compression
2.) g(x)=x^2-2
3.) j(x)=(x-1)^2
4.) h(x)=(x+2)^2+2
5.) g(x)=4/7x^2
6.) j(x)=(1/3x)^2
Use the description to write each quadratic function in vertex form.
7.) The parent function f(x)=x^2 is vertically stretched by a factor of 2.5 and translated 2 units left and 1 unit up to create h.
Using f(x)=x^2 as a guide, describe the transformations for each function.
8.) g(x)=8(x+2)^2
9.) p(x)=1/4x^2+2
10.) h(x)=-(1/3x)^2
11.) Which shows the fictions below in order from widest to narrowest of their corresponding graphs? a.) m,n,p,q b.) q,m,n,p c.) m,q,n,p d.) q,p,n,m

Transforming quadratic functions (need help SOON)?

Use the equation of motion:
s = ut + at^2 / 2
where s is the distance fallen, u the initial velocity, a the acceleration, and t the time.

With distances in ft, use g = 32 ft/sec^2.
The initial velocity u = 0.

Therefore:
s = 16t^2.

The height above ground is the initial height (400 or 1600 ft) minus the distance fallen.

(a)
h1(t) = 400 - 16t^2.
h2(t) = 1600 - 16t^2.

(b)
Start with t = 0, making h1(t) = 400 and h2(t) = 1600 as the first point on each of your two graphs. Then substitute increasing values for t and work out successive points. No matter what the value of t, it is clear that h2 is always 1200 greater than h1.

(c)
The sandbags reach the ground when h = 0.

h1 = 0 gives:
400 - 16t^2 = 0
400 = 16t^2
t^2 = 400 / 16
t = 20 / 4 = 5 sec.

h2 = 0 gives:
1600 - 16t^2 = 0
1600 = 16t^2
t^2 = 100
t = 10 sec.

Function transformations help? - 10 points?

1. to graph the function g(x)= x^3-9 start with the graph f(x) = x^3 and then make which changes?

shift the graph by what and which direction? reflect the graph over which axis? stretch the graph horizontally or vertically? (not all are necessary)

and

2. to graph the function g(x)= 3sqrt(-x-10) start with the graph f(x) = 3sqrt(x) and then make which changes?

shift the graph by what and which direction? reflect the graph over which axis? stretch the graph horizontally or vertically? (not all are necessary)

THANK YOU!

Write a quadratic function? Help!!?

Quadratic equations must include a variable term that has a 2 for an exponent. Without that exponent, then the equation becomes a linear equation, not a quadratic equation.

The basic quadratic equation is
y = ax^2 + bx + c
If the 'a' coefficient is positive, then the parabola opens upward.
If the 'a' coefficient is negative, then the parabola opens downward.
I will assume that it is positive so the parabola opens upward.
If we modify this equation to
y = a(x + d)^2 + bx + c
we still have a quadratic equation with four different control variables
'a' controls vertical shrinking and expanding
'b' controls downward with left shifting or upward with right shifting
'c' controls downward or upward shifting
'd' controls left or right shifting

For this problem, since the 'bx' term controls movement in two directions at the same time, I will eliminate it because I have all the movement control I need with the other three variables.
So the quadratic equation becomes:
y = a(x + d)^2 + c
This is still a quadratic equation because of the 2 exponent.

To cause the parabola to shift downward by 9, the 'c' term must be negative 9. A positive 9 would cause a shift upward by 9. So our equation changes to:
y = a(x + d)^2 - 9

Next, to shift the parabola 4 to the left, the 'd' term must be positive 4. A negative 4 would cause a shift to the right. So our equation changes to:
y = a(x + 4)^2 - 9

As mentioned earlier, when dealing with the 'a' coefficient, the parabola opens upward when this value is positive and the parabola opens downward when this value is negative. For this problem, I am assuming that it is positive and our parabola opens upward. So the 'a' coefficient will always be positive.
To vertically shrink by 3, the 'a' term must equal a positive 1/3. Multiplying the x term by 1/3 decreases the product by a factor of 3. If it was a positive 3, that would vertically expand by 3 because we would be multiplying the x term by 3.
So the quadratic equation for this problem becomes

y = (1/3)(x + 4)^2 - 9

/