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

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3 Answers

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 = (y2 - y1)/(x2 - x1)
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)
y = (11/4)x + b
-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)
y = (11/4)x + 4
4 = (11/4)(0) + 4
4 = 4
 
y = (11/4)x + 4
 
Determine the vertex of the parabola y = 4x2 - 80x + 406
  • We have to use the vertex formula x = -b/2a
  • where y = ax2 + bx + c
a = 4, b = -80
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 = 4x2 - 80x + 406
y = 4(10)2 - 80(10) + 406
y = 4(100) - 800 + 406
y = 400 - 394
y = 6
 
The vertex is (10,6)
 
Hi Brandi;
Find an equation of the line containing the pair of points (-4, -7) and (0, 4)
Our first priority is to establish slope.
Slope is change-of-y divided by change-of-x...
m=(y-y1)/(x-x1)
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).
y=(11/4)x+4
Let's use the other equation to verify our results...
-7=(11/4)(-4)+4
-7=-11+4
-7=-7

Determine the vertex of the parabola:
y= 4x^2-80x+406
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=4x2-80x+406
y=[(4)(102)]-[(80)(10)]+406
y=[(4)(100)]-[(80)(10)]+406
y=400-800+406
y=-400+406
y=6
The vertex is (10,6).
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
 
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!