can any one help me pls

Question

the point p (-1,4) lies on the curve with equation y=f(x).

state the coordinate of the image of point p on the curves with the following equation:

(i)y=f(x)+3

(ii)y=f(x+3)

Answer 1

Let's solve this step by step.

We are given the point P(−1,4)P(-1, 4)P(−1,4) on the curve y=f(x)y = f(x)y=f(x).

(i) Transformation: y=f(x)+3y = f(x) + 3y=f(x)+3

1. Adding 3 to the function shifts the graph vertically up by 3 units.
2. The xxx-coordinate stays the same; the yyy-coordinate increases by 3.

P(−1,4)→P′(−1,4+3)=(−1,7)P(-1, 4) \rightarrow P'(-1, 4 + 3) = (-1, 7)P(−1,4)→P′(−1,4+3)=(−1,7)

(ii) Transformation: y=f(x+3)y = f(x + 3)y=f(x+3)

1. Adding 3 inside the function argument shifts the graph horizontally to the left by 3 units.
2. The yyy-coordinate stays the same; the xxx-coordinate decreases by 3.

P(−1,4)→P′(−1−3,4)=(−4,4)P(-1, 4) \rightarrow P'(-1 - 3, 4) = (-4, 4)P(−1,4)→P′(−1−3,4)=(−4,4)

✅ Answer:

(i) (−1,7)(-1, 7)(−1,7)

(ii) (−4,4)(-4, 4)(−4,4)