Question

The graph of y=ax^2 + c contains the point (-3, 6). Which point lies on the graph of y=a(x+4)^2 + c?

A) (-7,6)
B) (1, 6)
C) (-3, 10)
D) (-3, 2)

Answer 1

The correct answer is option A, with the point being (-7,6), determined by accounting for the horizontal shift of (+4) in the new equation.

Step-by-step explanation:

The graph of y=ax^2 + c contains the point (-3, 6). We want to determine which point lies on the graph of y=a(x+4)^2 + c. If we take the given point (-3, 6) and apply it to the new equation, we need to adjust the x-coordinate according to the transformation represented by (x+4). To find the new coordinates, we set x + 4 equal to -3 so that we find the original x value that when increased by 4 results in -3. Solving the equation x + 4 = -3 gives us x = -7. Therefore, the new point that will have the same y-value of 6 on the transformed graph is (-7,6), which corresponds to option A.

Answer 2

The point that lies on the graph of
`\(y = a(x+4)^2 + c\)` given the point (-3, 6) on the graph of
`\(y = ax^2 + c\)` is point C) (-3, 10).

Step-by-step explanation:

The point (-3, 6) lies on the graph of
`\(y = ax^2 + c\)`. To find the corresponding point on the graph of
`\(y = a(x+4)^2 + c\)`, consider that shifting the point (-3, 6) four units to the left (adding 4 to the x-coordinate) will result in the point (-7, ?).

Since the y-coordinate is expected to remain the same due to the shifting in the x-coordinate, the y-value should still be 6. However, (-7, 6) is not among the provided options.

Alternatively, considering the vertex form of the quadratic equation
`\(y = a(x-h)^2 + k\)` where (h, k) represents the vertex of the parabola, the vertex is at (-4, c) in the equation
`\(y = a(x+4)^2 + c\)`. Given that (-3, 6) is on the graph of
`\(y = ax^2 + c\)`, if the point (-3, 6) is used to determine the value of 'c', it yields
`\(c = 6\)`.

Substituting c = 6 into
`\(y = a(x+4)^2 + c\)`, for x = -3:

`\(y = a(-3+4)^2 + 6\)`

`\(y = a(1)^2 + 6\)`

`\(y = a + 6\)`

Given that y = 6 when x = -3, a = 4. Therefore, when x = -3 in the equation
`\(y = a(x+4)^2 + c\)`:

`\(y = 4(-3+4)^2 + 6\)`

`\(y = 4(1)^2 + 6\)`

`\(y = 4 + 6\)`

`\(y = 10\)`

Hence, the point (-3, 10) lies on the graph of
`\(y = a(x+4)^2 + c\)`, which matches option C).