Math 2

Graph f(x) = ( x- 4)^2 - 2

4 units right of x axis and 2 units shifted down

y=- x^2+7

Negative parabola, facing down 7 units on y axis

Y= 3x^2

Parabola at the center facing up. Similar to y=x^2

F(x) = (x + 7)^4

Parabola is on x axis ,7 units to the left facing up

Function whose graph is the graph of y= x^3 + 2 but is reflected about x axis

- X^3 - 2

F(x) = x^2- 4x + 2

Calculate axis of symmetry. -b/2a. -_4x/2(1) = 1 .plug 1 in the main function to find the y intercept of the vertex. Plot the two point starting with x=2 . Y is -2 because vertex is (2,-2)

Graph Y= 1\2 X^2 + 1

Compress vertically by 1/2, there's no horizontal shift = 0 . Shift vertically up by one unit on y axis above (0,0) center of graph to (0,1)

Graph of f(x) = x^2 + 8 is

Shifting up Y axis 8 units

f(x) = x^5 - 2

Using transformation of y=x^5. The center point of cube function parabola on -2 y axis .

F(x) = 9/2x^4

Using transformation y = x^4, vertex is at (0,0) while sides of parabola are split in two equal halves of (1 and 9/2) besides y axis

function of graph y= x^3 shifted to the right 7 units

Y = ( x - 7)^3

Function of graph y= x^3 shifted down 6 units

Y= X^3 - 6

Function whose graph of Y= x^3 shifted down 8 units

Y=x^3- 8

Unit price is P dollars, revenue in dollars is R(p) = -6p^2 + 30000p . Unit price to maximize revenue? Find minimum revenue ?

-b/ 2a. - 30000/-6 = 2500 To get maximum revenue, plug 2500 in -6(2500)^2 + 30000(2500)

Function whose graph is graph of y= x^3 + 4 but is reflected on x axis

-x^3 - 4

Y= - |x+6|

Absolute value parabola facing down. 6 units left of x axis . _6

Graph F(x) = 3(x - 3)^2 - 5 starting with graph of y=x^2 and using transformations of shifting etc.

Compress vertically by 3:as in (a) Shift horizontally by 3 units right Shift vertically 5 units down on y axis

Graph f(x) = 2x^2 - x + 4. Graph opens up

Coordinates of vertex (1/4, 31/8). Axis of symmetry X= 1/4. Must be including X= then 1/4 Intercepts .use f (0) = 2(0)^2 - (0) + 4 to find (0, 4) as intercepts Graph will have 1/4 as axis of symmetry and will be shifted up by 3.9 (from 31/8) Domain is ( -&,&). Range [31/8, &) increase (1/4,&) . Decrease (-&, 1/4)

Quad f(x) = x^2 + 6x - 7. Find domain,range,increase and decrease

Domain (-&,&). Range [-16, &) Increase (-3,&), Decreasing at (-&, -3). Vertex -b/2a. -6/2*1. -6/2 = -3 is the axis of symmetry, it also determines increase and decrease

Graph f(x) = (x-6)^2 + 2

Looks like a chair facing left of x axis. Flat curve is on 2 units of y axis and 6 units of x axis, right at the center of flat curve. The parabola grpahl is 6 units on the right side of x axis, 2 units up y axis

F(x) = -2x^2 + 16x - 4 min or max value

Maximum value . Use -b/2a. Then plug. Pay attention to negative values.

F(x) = 2x^2 + 12x - 1 min or max value values

Minimum value . Use -b/2a. Then plug and find minimum value

Graph using techniques of shifting, etc y= 4 root x. Find domain and range

Reflect it under the x axis from the (0,0) vertex it arrow facing the right side of x axis Domain is (-&,&). Range is (-&,&)

Graph by shifting and compressing, reflecting, etc. from basic function y= root x. Graph h(x) = root x + 3 , domain? Range?

Shift function left from vertex (0,0) to -3 on the left side of x axis. Domain is (-&, &) Range [0, &)

Graph using techniques of shifting etc, g(x) = 3 root x-1 +2. Find domain and range

Shift graph up by negative 2 from (0,0) on y axis and one unit right from (0,0) on the x axis Domain in [1, &). Range is [2, &)

Graph function f(x) = (x+3)^2 - 1 starting with y = x^2 and using transformations of shifting

Shift parabola left by 3 units left, along x axis , and shift along y axis down by 1 unit from the -3 on x axis to y axis -1 value.

H(x) = -2x^5

Using transformations to graph the function. The cube function parabola flat curve is at (0,0) along x axis center. Looks like a chair facing right side of x axis.with vertex on (0,0)

Y= -3 |x| (neg 3 outside the absolute value x)

Vertex at x axis on -3 point, absolute value graph ,facing down