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Answer:
Both functions have one x-intercept each.
Step-by-step explanation:
The first function is
This is a parabola with vertices at the origin and has one x-intercept at t=0.
The transformed function is
The function g(t) is obtained by shifting the graph of f(t) to the left by 3 units.
This graph also has one x-intercept at x=-3.
Therefore both functions has the same number of x-intercepts
Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
recall that
therefore, let's just plug that on the remaining ones,
now, let's rationalize the denominator on tangent and secant,
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
recall that
therefore, let's just plug that on the remaining ones,
now, let's rationalize the denominator on tangent and secant,
Standard form for the equation of the line is
Solution:
Given point is (2, –5).
slope of the line m =
Here,
Equation of a line passing through the point:
Subtract 15 from both sides of the equation.
Standard form for the equation of the line is