Sketch a graph of f(x)= -2x+1

Mathematics

1 answer:

motikmotik3 years ago

7 0

Here is what you can sketch

You might be interested in

Show and/or explain that sin C = sin(A+B) in any & ABC.

Ne4ueva [31]

$\text{In any triangle ABC,}~\\\\A +B+C = \pi\\\\\text{Now,}\\\\\text{L.H.S}\\\\=\sin C\\\\=\sin\left[\pi - (A+B) \right]\\\\=\sin(A+B)\\\\=\text{R.H.S}\\\\\text{Showed.}$

4 0

3 years ago

Please help soon!

Jlenok [28]

Answer:

a) Translate the graph of f(x) down 3 units.

b) (8,-1) is a point on the graph of g(x)

Step-by-step explanation:

The transformations that subtract a constant number of units from the functional expression f(x) are transformations that lower the graph of the function in those many units,

Therefore they correspond to a translation of the original graph down the number of units involved.

In this case, the number of units involved is "3" (due to the "-3" added to the expression for f(x). So he correct answer for the first part is: Translate the graph of f(x) down 3 units.

For the second part, one has to try each of the coordinate pairs given in the new function g(x) to see which one results in a true statement:

1) Testing (-8,-1) by checking if replacing x with the value "-8" renders "-1" for the y-value: $g(x)=\sqrt[3]{x} -3\\g(-8)=\sqrt[3]{-8} -3\\g(-8)=-2-3\\g(-8)=-5$

so this is NOT a point on the graph of g(x).

2) Testing (-1,-2) by checking if replacing x with the value "-1" renders "-2" for the y-value: $g(x)=\sqrt[3]{x} -3\\g(-1)=\sqrt[3]{-1} -3\\g(-1)=-1-3\\g(-1)=-4$

so this is NOT a point on the graph of g(x).

3) Testing (2,-1) by checking if replacing x with the value "2" renders "-1" for the y-value: $g(x)=\sqrt[3]{x} -3\\g(2)=\sqrt[3]{2} -3\\$

the cubic root of 2 is not a rational number, because 2 is not a perfect cube, so the expression cannot be reduced, so this is NOT a point on the graph of g(x).

4) Testing (8,-1) by checking if replacing x with the value "8" renders "-1" for the y-value: $g(x)=\sqrt[3]{x} -3\\g(8)=\sqrt[3]{8} -3\\g(8)=2-3\\g(8)=-1$