How do I graph this
2 answers:
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Answer:
f(x) = -5x +48
Step-by-step explanation:
A fairly easy way to write the equation of the perpendicular line is to swap the x- and y-coefficients, negating one of them. Then choose the constant in the equation to make it true at the given point.
<h3>Swapped coefficients</h3>In this instance, we find it convenient to leave the coefficient of y as positive.
5y = x -10 . . . . . . original equation
y = -5x +c . . . . . . with coefficients swapped, x-coefficient negated
<h3>New constant</h3>At the given point, the equation becomes ...
8 = -5(8) +c
48 = c . . . . . . . add 40
<h3>Equation of the perpendicular line</h3>The equation of the line is then ...
y = -5x +48
f(x) = -5x +48 . . . . . . in functional notation
Answer:
Seee answer below.
Step-by-step explanation:
a. k = −1
If K=-1 the equation gets this form:
(x^2/-1) -y^2=1
There aren't natural numbers that being negative, adding them, we get 1 as result. So there is no graph for this equation.
b. k = 1
(x^2/1) -y^2=1
This is the natural form of the equation of an hyperbola. Attached you can find the graph.
c. k = 2
(x^2/2) -y^2=1
This is the natural form of the equation of an hyperbola. Attached you can find the graph.
d. k = 4
(x^2/4) -y^2=1
This is the natural form of the equation of an hyperbola. Attached you can find the graph.
e. k = 10
(x^2/10) -y^2=1
This is the natural form of the equation of an hyperbola. Attached you can find the graph.
f. k = 25
(x^2/25) -y^2=1
This is the natural form of the equation of an hyperbola. Attached you can find the graph.
g. Describe what happens to the graph of
x2 / k − y2 = 1 as k → [infinity].
As K is increasing the value of X will be tending to 0. So the equation for this will be:
− y^2 = 1.The solution for this is in the domain of the imaginary numbers.
