if the function f(x) has a domain of -2≤x≤8 and a range of -4≤y≤6 and the function g(x) is defined by the formula g(x)= 5f(2x) t
1 answer:
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Answer:
Option A: b must equal 7 and a second solution to the system must be located at the point (2, 5)
Step-by-step explanation:
<u><em>The complete question is</em></u>
Tom determines that the system of equations below has two solutions, one of which is located at the vertex of the parabola.
Equation 1: (x – 3)2 = y – 4
Equation 2: y = -x + b
In order for Tom’s thinking to be correct, which qualifications must be met?
A: b must equal 7 and a second solution to the system must be located at the point (2, 5).
B: b must equal 1 and a second solution to the system must be located at the point (4, 5).
C: b must equal 7 and a second solution to the system must be located at the point (1, 8).
D: b must equal 1 and a second solution to the system must be located at the point (3, 4).
step 1
Find the vertex of the quadratic equation
The general equation of a vertical parabola in vertex form is
where
(h,k) is the vertex
we have
so
The vertex is the point (3,4)
step 2
Find out the value of b in the linear equation
we know that
If the vertex is a solution of the system of equations, then the vertex must satisfy both equations
substitute the value of x and the value of y of the vertex in the linear equation
For x=3, y=4
so
step 3
Find out the second solution of the system of equations
we have
-----> equation A
----> equation B
solve the system of equations by graphing
Remember that the solutions are the intersection points both graphs
The second solution of the system of equations is (2,5)
see the attached figure
therefore
b must equal 7 and a second solution to the system must be located at the point (2, 5)
Answer:
5x -7y = 21
Step-by-step explanation:
A sketch can convince you that BC is a transversal perpendicular to parallel lines AB and CD. The question asks for an equation for CD, so we just need to write the equation of a line through D that is parallel to AB.
One way to do this is to equate the slopes of the parallel lines:
∆y/∆x for AB = ∆y/∆x for CD
(y2 -y1)/(x2 -x1) = (y -2)/(x -7) . . . . . . where (x1, y1) = A; (x2, y2) = B; (7, 2) = D
(4 -(-1))/(1 -(-6)) = (y -2)/(x -7)
5(x -7) = 7(y -2) . . . . . . . . . . . . . cross multiply
5x -7y = 21 . . . . . . . . . . . . . . . . add 35 -7y, simplify
_____
Note that the graph shows the line CD is named "b", and its equation is shown at upper left. Multiplying that equation by -1 gives the one shown here.
