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Answer:
1.
<u>An extraneous solution is a root of a transformed equation that is not a root of the original equation as it was excluded from the domain of the original equation.</u>
It emerges from the process of solving the problem as a equation.
2.I begin like:
The vertical asymptotes will occur at those values of x for which the denominator is equal to zero:
for example:
x² − 4=0
x²= 4
doing square root on both side
x = ±2
Thus, the graph will have vertical asymptotes at x = 2 and x = −2.
To find the horizontal asymptote, the degree of the numerator is one and the degree of the denominator is two.
Answer:
a. The slope of a line parallel to the given line is 1
b. A point on the line parallel to the given line, passing through (−4, 2), is (1,7)
c. The slope of the line perpendicular to the given line is -1
d. A point on the line perpendicular to the given line, passing through (−4, 2), is (3,-5)
Step-by-step explanation:
The equation of the line in Slope-intercept form is:
Where m is the slope and b is the y-intercept.
a. For the line
You can identify that:
By definition, two lines are parallel if they have the same slope. Then, the slope of a line parallel to the given line is:
b. The equation of the line in Point-slope form is:
Where m is the slope and () is a point of the line.
Given the point (-4,2), substitute this point and the slope of the line into the equation:
Give a value to "x", substitute it into this equation and solve for "y":
For :
Then, you get the point (1,7)
c. The slopes of perpendicular lines are negative reciprocals, then the slope of a line perpendicular to the given line is:
d. Given the point (-4,2), substitute this point and the slope of the line into the equation:
Give a value to "x", substitute it into this equation and solve for "y":
For :
Then, you get the point (3,-5)