Find an equation of the line containing the pair of points (-4, -7) and (0, 4)

Determine the vertex of the parabola:

y= 4x^2-80x+406

Find an equation of the line containing the pair of points (-4, -7) and (0, 4)

Determine the vertex of the parabola:

y= 4x^2-80x+406

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Broomfield, CO

Find an equation of the line containing the pair of points (-4,-7) and (0,4)

- With no further information I take it this is a linear equation
- y = mx + b
- First we have to find the slope (m)

m = (4 - (-7))/(0 - (-4))

m = (4 + 7)/(0 + 4)

m = 11/4

- Next we have to solve for the y-intercept (b)
- We'll use the first set to solve (-4,-7)

-7 = (11/4)(-4) + b

-7 = -11 + b

-7 + 11 = b

b = 4

- Now we will use the second set to check our work (0,4)

4 = (11/4)(0) + 4

4 = 4 √

Determine the vertex of the parabola y = 4x^{2} - 80x + 406

- We have to use the vertex formula x = -b/2a
- where y = ax
^{2}+ bx + c

x = -b/2a

x = -(-80)/2(4)

x = 80/8

x = 10

- Now we use this value of x in the original equation to solve for y

y = 4(10)^{2} - 80(10) + 406

y = 4(100) - 800 + 406

y = 400 - 394

y = 6

Middletown, CT

Hi Brandi;

Our first priority is to establish slope.

Slope is change-of-y divided by change-of-x...

m=(y-y_{1})/(x-x_{1})

m=(-7-4)/(-4-0)

m=-11/-4

A negative number divided by a negative number has a positive result...

m=11/4

The equation in slope-intercept form is...

y=mx+b

m is the slope.

b is the y-intercept, the value of y when x=0.

The slope is (11/4).

In this circumstance, the y-intercept is provided as (0,4).

Let's use the other equation to verify our results...

-7=(11/4)(-4)+4

-7=-11+4

-7=-7

The vertex is the value of x at the point of the parabola in which the change of slope is zero. Henceforth, we will take the first derivative of the equation and set this equal to zero...

0=8x-80

80=8x

10=x

Let's establish the value of y...

y=4x^{2}-80x+406

y=[(4)(10^{2})]-[(80)(10)]+406

y=[(4)(100)]-[(80)(10)]+406

y=400-800+406

y=-400+406

y=6

The x value of the vertex can also be established by applying the equation -b/2a to the original formula.

y= 4x^2-80x+406

-b/2a

x=-[-80/(2)(4)]

x=80/8

x=10

Westford, MA

Find an equation of the line containing the pair of points (-4,-7) and (0,4)

y - -7 = ((-7-4)/(-4-0))(x - -4)

or

y - 4 = ((4 - -7)/(0 - -4))(x - 0)

=====

Determine the vertex of the parabola:

y = 4x^2 - 80x + 406

If the vertex is (h,k), then

h = -b/(2a) = -(-80)/(2(4)) = 10, and

k = y(h) = 4(10)^2 - 80(10) + 406

= 400 - 800 + 406

k = 6

Vertex is (10,6).

check:

4(x - 10)^2 + 6 =? 4x^2 - 80x + 406

4(x^2 - 20x + 100) + 6 =? 4x^2 - 80x + 406

4x^2 - 80x + 400 + 6 = 4x^2 - 80x + 406 checked!

y - -7 = ((-7-4)/(-4-0))(x - -4)

or

y - 4 = ((4 - -7)/(0 - -4))(x - 0)

=====

Determine the vertex of the parabola:

y = 4x^2 - 80x + 406

If the vertex is (h,k), then

h = -b/(2a) = -(-80)/(2(4)) = 10, and

k = y(h) = 4(10)^2 - 80(10) + 406

= 400 - 800 + 406

k = 6

Vertex is (10,6).

check:

4(x - 10)^2 + 6 =? 4x^2 - 80x + 406

4(x^2 - 20x + 100) + 6 =? 4x^2 - 80x + 406

4x^2 - 80x + 400 + 6 = 4x^2 - 80x + 406 checked!

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