Plot the points on the graph of f(x)=1/2(x+6)^2-5 that corresponds to x -values of -8 and -10 then plot the vertex and the reflections of these points in the axis of symmetry

Answer 1

This is an equation with a power of x², a parabola.

f(x) = ¹/₂ (x + 6)² - 5

We have to plot the points and draw the curve.

I think it's also helpful to know the vertex form of the parabola, y = a(x – h)² + k with the vertex at (h, k). But it's not strictly necessary here in order to find the points asked for.

for x = -8

y = ¹/₂ (-8 + 6)² - 5 = ¹/₂ (-2)² - 5 = -3

(-8, -3)

for x = -10

y = ¹/₂ (-10 + 6)² - 5 = ¹/₂ (-4)² - 5 = 3

(-10, 3)

the x term is zero when x = -6

y = ¹/₂ (-6 + 6)² - 5 = ¹/₂ (0)² - 5 = -5

vertex is at (-6, -5)

The parabola is symmetrical with the axis of symmetry passing through the vertex. This line is vertical with equation x = -6. You might find it helpful to draw it.

If the axis of symmetry is vertical (x = something) then the y value of the reflected point remains unchanged, only the x value changes.

for (-10, 3) it is 4 units to the left of the axis (x = -6), so the reflected point will be 4 units to the right (x=-2), so the point is (-2,3).

for (-8, -3) it is 2 units to the left of the axis (x = -6), so the reflected point will be 2 units to the right (x=-4), so the point is (-4, -3).

You now have the axis of symmetry, the vertex, and two additional points on each side of the curve. You should be able to draw a nice parabola curve now.