Absolute and Local max and min values of F

Question

Sketch the graph of f by hand and use your sketch to find the absolute and local maximum and minimum values of f.f (If an answer does not exist, enter DNE.)

f(x) = 3 + (x + 1)²

Answer 1

Explanation without Calculus:

This is going to be an upward facing parabola. The (x+1) inside the square shifts it 1 to the left because try plugging in 0, that's going to make it 0+1 = 1 inside the square, so at zero the graph is going to look like a non-shifted parabola would look like at 1, aka the 1 value is shifted over to the left. (when i first learned this stuff i was never taught it like that and just told to remember if it says 'plus one' its shifted left and 'minus one' would be shifted to the right.

Also, since there's a +3 on its own, if the whole squared part was zero, then y=3, but on a normal non-shifted parabola like y=x² it would just be zero when the x term is zero, so this graph is shifted 3 up (so now that's one to the left and three up for shifting the vertex).

Notice at x = -1, y = 3, and this is the minimum value of the graph, because adding any x value would increase the y value. Even negative x values would because it gets squared so that makes it positive and you get y = 3+some positive value, so this point (-1,3) is a minimum.

For calculus:

f'(x) = 2x+2 and to find the minimum you set it equal to zero so you get there's a minimum at x = -1, you can show the 2nd derivative is positive everywhere (it just equals 2) to show that the parabola is facing upwards. Plug in x =-1 and get 3 so the minimum is (-1,3) and you can get an easy sketch of the parabola by doing that... if you wanna be more exact and find the y intercept set x = 0, and get 4... since it's a parabola it's symmetric so at x = -2 the value is 4 as well (or just plug in -2)