6. In Problems i-iv, find each of the following: a) Intercepts b) Vertex c) Maximum or minimum d) Range

Question

i) m(x) = (𝑥+1)2 – 2 ii) n(x) = (𝑥−4)2 – 3

please solve, thank you very much in advance.

Answer 1

i.

Start with the vertex, since it's in vertex form. Vertex form is y = a(x - h)² + k where the vertex is at (h, k).

Therefore, we know the vertex of m(x) = (x + 1)² - 2 is (-1, -2).

b: (-1, -2)

We know that the y-intercept occurs where x = 0 and the x-intercepts occur where y = 0, so we can plug those values into the equation and solve.

m(x) = (x + 1)² - 2

m(0) = (0 + 1)² - 2

m(0) = -1

So, the y-intercept is at (0, -1).

0 = (x + 1)² - 2

0 = (x + 1)² - 2

0 = x² + 2x - 1

This is not easily factorable, so we can use the quadratic formula:

( -b +- √b² - 4ac ) / (2a) - (-2 +- √2²-4(1)(-1) ) / 2(1)

And by solving, we see that the two x-intercepts are at approximately (0.41, 0) and (-2.41, 0).

a: (0, -1) (0.41, 0) (-2.41, 0)

If we plot all the points we've drawn so far, we can see that we have a parabola facing upwards, meaning that the minimum point on the graph is at the vertex, (-1, -2).

c: minimum @ (-1, -2)

The y value at the vertex (-3) is the lowest the graph goes, but it extends upwards infinitely, meaning that the range is from -3 (inclusive) to infinity.

d: [-3, ∞)

ii.

We can use very much the same logic for the second equation and we get:

a: (4, -3)

b: (0, 13) (2.27, 0) (5.73, 0)

c: minimum @ (4, -3)

d: [-3, ∞)

Upside-down parabolas will have a negative "a" term. For example,

y = -2(x + 3)² + 2

Where a = -2

They will also have maximums at their vertexes and extend downward infinitely, so this graph would have a maximum at (-3, 2) and a range of (-∞, 2].