Question

can some one help me please
igreen

Answer 1

Create a quadratic function, f(x), in vertex form. The a should be between 4 and –4, the h will be your birth month, and the k will be your birth day. Write your equation below.

my birth month is october and birthday 6

Answer 2

@gorv @iGreen

Answer 3

@kfullwood1

Answer 4

hey

Answer 5

Okay, vertex form of a function is: \( f (x) = a(x - h)^2 + k\).

Answer 6

ok so that would be solution problem

Answer 7

h will be birth month (10)

k will be birth day (6)

Answer 8

f(x)=a(x-10)^2+6

Answer 9

\(f (x) = a(x - h)^2 + k\)

We can pick 2 for 'a'.

So we have:

\(f (x) = 2(x - 10)^2 + 6\)

Answer 10

Yes, you got it.

Answer 11

That's your equation right there. Dont' forget the 2 before the parenthesis.

Answer 12

thx could i get help with 1 more its kind spin of of this problem

Answer 13

Using complete sentences, explain how to convert your birthday function into standard form.
the function is f(x)=a(x-10)^2+6

Answer 14

The function is f(x) = 2(x-10)^2 + 6.

Answer 15

Standard form is \(Ax + By = C\) I believe.

Answer 16

ok thank yah ...but how did yah get the standard form

Answer 17

btw sry for all questions

Answer 18

sry just finished helping someone in history

Answer 19

Oh, wait I think standard form is \(y = ax^2 + bx + c\)

Answer 20

oh ok

Answer 21

thxXD

Answer 22

I'm not sure..your lesson should tell you how to convert it..can you find that in there for me?

Answer 23

It should tell you how to convert "Vertex Form to Standard Form".

Answer 24

kk one sec ill search

Answer 25

y=(x−5)2−4
Identify the vertex, equation for the axis of symmetry, domain, range and x- and y-intercepts of the parabola.

Check My Work
From the general form of the parabola, y=a(x−h)2+k, the vertex is given by (h,k). So in this case, the vertex is (5,−4).

The axis of symmetry is found by setting x−5=0, giving you x=5.

Because any real number can be substituted for x in the function, the domain is all real numbers.

Since a is positive, the parabola opens up. The minimum value of the parabola occurs at y=−4; therefore the range is all real numbers greater than or equal to −4.

The x-intercepts occur where y=0 and solving for x. Because the solutions to 0=(x−5)2−4 are x=3 and x=7, the x-intercepts are (3,0) and (7,0). The y-intercept occurs where x=0 and solving for y. Because (0−5)2−4=21, the y-intercept is (0,21).

Answer 26

A quadratic equation is an equation of degree 2. The graph of a quadratic equation is in the shape of a parabola which looks like an arc. The general form of the equation is represented by

f(x) = a(x – h)2 + k

The vertex, or turning point, of the parabola is an ordered pair represented by (h, k). The vertex of the quadratic equation f(x) = –(x – 2)2 + 9 is (2, 9).

Using the average rate of change, it is possible to find the slope between two points and observe whether the graph’s rate of change is increasing or decreasing.

The axis of symmetry is a line that splits the parabola in half so that both halves are symmetrical to one another. The equation for this line is found by setting the expression inside the parentheses equal to 0 and solving for x.

f(x) = –(x – 2)2 + 9

x – 2 = 0

+2 _+2

x = 2

The domain of the quadratic equation is the set of numbers that can be substituted for x and result in a unique value for y. In the case of the equation shown here, the domain is "all real numbers."

The range of the quadratic equation is the set of numbers that are produced from the domain values of x. Algebraically, determine whether the vertex is the minimum or maximum point of the graph. If a is positive, the vertex is the minimum point, and the range is all of the y-coordinates greater than or equal to the y-coordinate of the vertex. If a is negative, the vertex is the maximum point, and the range is all of the y-coordinates less than or equal to the y-coordinate of the vertex.

a graph of a parabola opening down passing through points ( 0, 5 ), (5, 0), and ( 2, 9).

Graphically, it is easy to see the vertex is a maximum point and y-coordinates greater than 9 do not have a corresponding x-coordinate. Therefore, in the case of the equation shown, the range is y ≤ 9.

The intercepts of a quadratic equation are the places on the graph where the parabola crosses the x- or y-axis. The x-intercepts are found by looking at the graph of the parabola where it crosses the x-axis.

The standard form of the quadratic equation is represented by

f(x) = ax2 + bx + c

While the domain, range and intercepts of the parabola may be found in the same way as the general form of the parabola, the axis of symmetry and vertex must be found using the formula

x equals negative b over 2a
The axis of symmetry is found first using this equation.

f(x) = x2 + 6x + 8
x equals negative 6 over 2 times 1

x equals negative 6 over 2

x = –3

Then the vertex is found by substituting the x-coordinate of –3 in the original equation and solving for y to find the y-coordinate.

f(x) = x2 + 6x + 8
f(x) = (–3)2 + 6(–3) + 8
f(x) = 9 + 6(–3) + 8
f(x) = 9 – 18 + 8

f(x) = –1 (–3, –1)

Answer 27

Okay, this helps.

Answer 28

np

Answer 29

should i close the question or yah still looking

Answer 30

btw i have another one if you want to help with that 0,o

Answer 31

Hold on..

Answer 32

kk

Answer 33

Okay...I don't really have any idea how to do that..xD Sorry.

Answer 34

What's your other question?

Answer 35

4. Using complete sentences, explain how to find the average rate of change for f(x) from x = 4 to x = 7.

Answer 36

its ok if yah dont know how to do it

Answer 37

Plug in x = 4 and x = 7:

\(x = 4\)

\(f(x) = 2(x-10)^2 + 6\)

\(f(x) = 2(4-10)^2 + 6\)

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

\(f(x) = 2(36) + 6\)

\(f(x) = 72 + 6\)

\(f(x) = 78\)

So our first point is (4, 78).

\(x = 7\)

\(f(x) = 2(x-10)^2 + 6\)

\(f(x) = 2(7-10)^2 + 6\)

\(f(x) = 2(-3)^2 + 6\)

\(f(x) = 2(9) + 6\)

\(f(x) = 18 + 6\)

\(f(x) = 24\)

So our second point is (7, 24).

Answer 38

Now that we have our two points, (4, 78) and (7, 24), we can plug them into the average rate of change formula(slope formula), \(m = \dfrac{y_2-y_1}{x_2-x_1}\).

Answer 39

one sec let me go over what yah saying

Answer 40

oh ok i get it

Answer 41

so that would be solution thx

Answer 42

\(m = \dfrac{y_2-y_1}{x_2-x_1}\)

Plug in (4, 78) and (7, 24)

\(m = \dfrac{24-78}{7-4}\)

\(m = \dfrac{-54}{~~~3}\)

\(m = -18\)

So the average rate of change between x = 4 and x = 7 is -18.

Answer 43

:) ttyl