Compare the line passing through the points (−3, −11) and (6, 4) to the line given by the equation y=35x−6.

Mathematics

1 answer:

Aleks04 [339]3 years ago

7 0

Let
A------> (−3, −11)
B------> (6, 4)

step 1
find the slope m point A and point B
m=(y2-y1)/(x2-x1)-----> m=(4+11)/(6+3)----> m=15/9---> 5/3

step 2
with m=5/3 and point B (6, 4) find the equation of a line
y-y1=m*(x-x1)-----> y-4=(5/3)*(x-6)---> y=(5/3)x-6

case a. they have the same slope------> is not correct
because
y=35x−6------> the slope m=35
y=(5/3)x-6---> the slope m =5/3

case b. they have the same x-intercept.----> is not correct
because
the x-intercept is for y=0

y=35x−6-----> 0=35x-6----> 35x=6-----> x=6/35
the x-intercept is the point (6/35,0)

y=(5/3)x-6----> 0=(5/3)x-6----> (5/3)x=6---> x=18/5
the x-intercept is the point (18/5,0)

case c. the two lines are perpendicular. do. they have the same y-intercept-----> is nor correct
because

if two lines are perpendicular
m1*m2=-1-------> this condition is not satisfied

the y intercept is for x=0

y=35x−6-------> y=-6
the y intercept is the point (0,-6)

y=(5/3)x-6-----> y=-6
the y intercept is the point (0,-6)

they have the same y-intercept but the two lines are not perpendicular

see the attached figure

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Consider the parabola given by the equation: f(x) = 4x² - 6x - 8 Find the following for this parabola: A) The vertex: Preview B)

jeyben [28]

Answer:

The vertex: $(\frac{3}{4},-\frac{41}{4} )$

The vertical intercept is: $y=-8$

The coordinates of the two intercepts of the parabola are $(\frac{3+\sqrt{41} }{4} , 0)$ and $(\frac{3-\sqrt{41} }{4} , 0)$

Step-by-step explanation:

To find the vertex of the parabola $4x^2-6x-8$ you need to:

1. Find the coefficients a, b, and c of the parabola equation

$a=4, b=-6, \:and \:c=-8$

2. You can apply this formula to find x-coordinate of the vertex

$x=-\frac{b}{2a}$ , so

$x=-\frac{-6}{2\cdot 4}\\x=\frac{3}{4}$

3. To find the y-coordinate of the vertex you use the parabola equation and x-coordinate of the vertex ( $f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c$ )