What are the coordinates of the y-intercept of the graph of y=–4x+5? (0, 5) (0, -5) (-5, 0) (5, 0)

(0, 5)

Step-by-step explanation:

y = mx + b <== slope-intercept form of a straight line

m <== slope of the line

b = y-intercept (when x = 0)

y = -4x + 5

Therefore, for this line, the y-intercpet is (0,5)

Hope this helps!

-5

Step-by-step explanation

Given the function  f(x)=-4x^2+2x-5, the y intercept occurs at x = 0

Substitute x = 0 into the function and find y as shown;

f(x)=-4x^2+2x-5

f(0)=-(0)^2+2(0)-5

f(0)=0-5

f(0) = -5

The coordinate point is (0, -5)

Hence the value of the y coordinated is -5

say you had an equation y=5x+17 the 17 is the y intercept or where the line comes off the y-axis and the 5x is the slope or the rise over run 5/1 is the slope so going up from the y intercept would be 1 up and 5 over and thats how you would graph it.

say you had an equation y=5x+17 the 17 is the y intercept or where the line comes off the y-axis and the 5x is the slope or the rise over run 5/1 is the slope so going up from the y intercept would be 1 up and 5 over and thats how you would graph it.

a) (-5,0) and (1,0)

b) (0,-5)

c) minimum

See attached graph.

Step-by-step explanation:

To graph the function, find the vertex of the function find (-b/2a, f(-b/2a)). Substitute b = 4 and a = 1.

-4/2(1) = -4/2 = -2

f(-2) = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -4 - 5 = -9

Plot the point (-2,-9). Then two points two points on either side like x = -1 and x = -3. Substitute x = -1 and x = -3

f(-1) = (-1)^2 + 4 (-1) - 5 = 1 - 4 - 5 = -8

Plot the point (-1,-8).

f(-3) = (-3)^2 + 4(-3)  - 5 = 9 - 12 - 5 = -8

Plot the point (-3,-8).

See the attached graph.

The features of the graph are:

a) (-5,0) and (1,0)

b) (0,-5)

c) minimum

a) (-5,0) and (1,0)

b) (0,-5)

c) minimum

See attached graph.

Step-by-step explanation:

To graph the function, find the vertex of the function find (-b/2a, f(-b/2a)). Substitute b = 4 and a = 1.

-4/2(1) = -4/2 = -2

f(-2) = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -4 - 5 = -9

Plot the point (-2,-9). Then two points two points on either side like x = -1 and x = -3. Substitute x = -1 and x = -3

f(-1) = (-1)^2 + 4 (-1) - 5 = 1 - 4 - 5 = -8

Plot the point (-1,-8).

f(-3) = (-3)^2 + 4(-3)  - 5 = 9 - 12 - 5 = -8

Plot the point (-3,-8).

See the attached graph.

The features of the graph are:

a) (-5,0) and (1,0)

b) (0,-5)

c) minimum

1) The equation is given in factored form

2) The key characteristics of the parabola are;

a. The parabola has 2 real roots and extends to infinity on both sides

b. The x-intercepts of the parabola are (1/2, 0) and (-6, 0)

c. The y intercept is (0, -6)

d. The vertex point is (-2.75, -21.125)

Step-by-step explanation:

The given equation is f(x) = (2·x -1)(x + 6)

Therefore the equation is given in factored form

The key characteristic of the parabola revealed from the form f(x) = (2·x -1)(x + 6) are;

1) Writing the parabola in the intercept form, a(x - p)(x -q) we have;

f(x) = (2·x -1)(x + 6) = 2·(x - 1/2)(x + 6)

p = 1/2, q = -6

Therefore;

Given that a is positive the parabola is concave upwards

2) a. The parabola has 2 real roots and extends to infinity on both sides

b. The x-intercepts of the parabola are p and q which are (1/2, 0) and (-6, 0)

c. The y intercept is a·(-p)·(-q) = 2×(-1/2)×6 = -6

The y intercept is (0, -6)

d. Expanding we get;

f(x) = 2·x² + 11·x - 6

From the

The vertex is h = -b/(2·a) = -11/(2×2) = -2.75

h(2.75) = 2×(-2.75)² + 11×(-2.75) - 6 = -21.125

Therefore, the function is symmetrical about the vertex point (-2.75, -21.125)

1) The equation is given in factored form

2) The key characteristics of the parabola are;

a. The parabola has 2 real roots and extends to infinity on both sides

b. The x-intercepts of the parabola are (1/2, 0) and (-6, 0)

c. The y intercept is (0, -6)

d. The vertex point is (-2.75, -21.125)

Step-by-step explanation:

The given equation is f(x) = (2·x -1)(x + 6)

Therefore the equation is given in factored form

The key characteristic of the parabola revealed from the form f(x) = (2·x -1)(x + 6) are;

1) Writing the parabola in the intercept form, a(x - p)(x -q) we have;

f(x) = (2·x -1)(x + 6) = 2·(x - 1/2)(x + 6)

p = 1/2, q = -6

Therefore;

Given that a is positive the parabola is concave upwards

2) a. The parabola has 2 real roots and extends to infinity on both sides

b. The x-intercepts of the parabola are p and q which are (1/2, 0) and (-6, 0)

c. The y intercept is a·(-p)·(-q) = 2×(-1/2)×6 = -6

The y intercept is (0, -6)

d. Expanding we get;

f(x) = 2·x² + 11·x - 6

From the

The vertex is h = -b/(2·a) = -11/(2×2) = -2.75

h(2.75) = 2×(-2.75)² + 11×(-2.75) - 6 = -21.125

Therefore, the function is symmetrical about the vertex point (-2.75, -21.125)

The answer is in the image