C. The variable x has a coefficient. B. (2, 10) C. The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is x=0. The maximum value of the function is 25. D. The student is not correct. The equation f(x)=0 does not have any real solutions, so the graph does not have any x-intercepts.

Step-by-step explanation:

1) Why is f(x)=(3x+1/3)^2+8/9 not the vertex form of f(x)=9x^2+2x+1?

A.The expression has a constant outside of the squared term.

B. The expression is not the product of two binomials.

C. The variable x has a coefficient.

D. Some of the terms are fractions instead of integers.

Vertex form is a(x -h)^2 +k. The coefficient of x inside parentheses is 1. The given form is not vertex form because the leading coefficient has not been removed to outside parentheses.

__

2) What is the vertex of the parabola with the equation y=(x−2)^2+10?

A. (−2, −10)

B. (2, 10)

C. (−2, 10)

D. (2, −10)

Vertex form is a(x -h)^2 +k. Comparing to the given equation, we find the vertex (h, k) = (2, 10).

__

3) For the given function, identify the x- and y-intercepts if any, the vertex, the axis of symmetry, and the maximum or minimum value. f(x)=−x^2+25

A. The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,−25). The vertex is (0,−25). The axis of symmetry is x=0. The minimum value of the function is −25.

B. There are no x-intercepts. The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is y=0. The maximum value of the function is 25.

C. The x-intercepts are (−5,0) and (5,0). The y-intercept is (0,25). The vertex is (0,25). The axis of symmetry is x=0. The maximum value of the function is 25.

D. The x-intercepts are (−25,0) and (25,0). The y-intercept is (0,5). The vertex is (0,5). The axis of symmetry is x=0. The maximum value of the function is 5.

The x-intercepts are the values of x that make y=0. They are (±5, 0). The y-intercept is the value of y when x=0. It is (0, 25).

__

4) A student says that the function f(x)=−x^2−9 has the x-intercepts (−3,0) and (3,0). Is the student correct? If not, explain why.

A. The student is correct.

B. The student is not correct. The equation f(x)=0 has one real solution, so the x-intercept is (9,0).

C. The student is not correct. The equation f(x)=0 does not have any real solutions, so the graph has only one x-intercept, (0,0).

D. The student is not correct. The equation f(x)=0 does not have any real solutions, so the graph does not have any x-intercepts.

The parabola opens downward and has a maximum value of -9, so cannot cross the x-axis. There are no x-intercepts, hence no real solutions.

The first equation: x= -5/4,0 y= 0,-5

The second equation: y=2

The third equation: is the same as the first one

Step-by-step explanation:

Part 1)

we know that

the equation of the line in slope-intercept form is equal to

where

m is the slope

b is the y-intercept

we have

solve for y

-------> equation of the line in slope-intercept form

so

the slope m is

the y-intercept b is

Part 2)

we know that

the equation of the line in slope-intercept form is equal to

where

m is the slope

b is the y-intercept

we have

solve for y

-------> equation of the line in slope-intercept form

so

the slope m is

the y-intercept b is

Part 3)

we know that

The x-intercept is the value of x when the value of y is equal to zero

The y-intercept is the value of y when the value of x is equal to zero

we have

a) Find the x-intercept

For substitute in the equation

The answer part 3a) is

b) Find the y-intercept

For substitute in the equation

The answer part 3b) is

Part 4)

we know that

the equation of the line in standard form is

we have

Multiply by both sides

------> equation in standard form

therefore

the answer Part 4) is option B False

Part 5)

Step 1

Find the slope

we have

solve for y

so

the slope m is

Step 2

Find the y-intercept

The y-intercept is the value of y when the value of x is equal to zero

we have

for

the y-intercept is

Step 3

Find the equation of the line

we have

the equation of the line in slope-intercept form is

substitute the values

therefore

the answer Part 5) is the option A

Part 6)

Step 1

Find the slope of the given line

we know that

if two lines are perpendicular. then the product of their slopes is equal to minus one

so

in this problem

the given line

solve for y

the slope m1 is

so

the slope m2 is

Step 2

Find the equation of the line

we know that

the equation of the line in slope point form is equal to

we have

point

substitutes the values

therefore

the answer part 6) is the option C

Part 7)

-------> the slope is

--------> the slope is

we know that

if two lines are parallel , then their slopes are the same

in this problem the slopes are not the same

therefore

the answer part 7) is the option D) No, since the slopes are different.

Part 8)

a. Write an equation for the line in point-slope form

b. Rewrite the equation in standard form using integers

Step 1

Find the slope of the line

we know that

the slope between two points is equal to

substitute the values

Step 2

Find the equation in point slope form

we know that

the equation of the line in slope point form is equal to

we have

point

substitutes the values

-------> equation of the line in point slope form

Step 3

Rewrite the equation in standard form using integers

Multiply by both sides

--------> equation of the line in standard form

Part 9)

we know that

The formula to calculate the slope between two points is equal to

where

(x1,y1) ------> is the first point

(x2,y2) -----> is the second point

In the numerator calculate the difference of the y-coordinates

in the denominator calculate the difference of the x-coordinates

Part 10)

we know that

The formula to calculate the slope between two points is equal to

substitutes

therefore

the answer Part 10) is

Part 11)

we know that

the equation of the line in slope point form is equal to

substitute the values

--------> this is the equation in the point slope form

1 ) They are perpendicular.

2 ) m · (-1) / m = -1.

The product of the slopes is -1.

3 ) The center of the circle is ( 2, 2 ).

4 ) m = (6-2 ) / 5 -2 = 4/3

5 ) Slope of the tangent: m = - 3/4.

6 ) m = -3/4, passes through the point: ( 5. 6 ):

6 = - 15/4 + b

b = 24/4 + 15/4

b = 39/4

The slope-intercept equation is:

y = -3/4 x + 39/4

7) We will put : x = 0 in the linear equation.

8 ) We will put y = 0 in the linear equation.

9 ) y-intercept : y = 9.75

Zero: x = 13.

2 ) m · (-1) / m = -1.

The product of the slopes is -1.

3 ) The center of the circle is ( 2, 2 ).

4 ) m = (6-2 ) / 5 -2 = 4/3

5 ) Slope of the tangent: m = - 3/4.

6 ) m = -3/4, passes through the point: ( 5. 6 ):

6 = - 15/4 + b

b = 24/4 + 15/4

b = 39/4

The slope-intercept equation is:

y = -3/4 x + 39/4

7) We will put : x = 0 in the linear equation.

8 ) We will put y = 0 in the linear equation.

9 ) y-intercept : y = 9.75

Zero: x = 13.

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