Question down there to big for up here.....?

Question

Find the vertex of the parabola y=3(x+1)^2-4. Give your answer by filling in the blacks in the following sentence, where the first blank is the coordinates or the vertex, and the second blank is either the word "highest" or the word "lowest".

The Vertex is ___________, and this point is the ____________ point on the parabola.

Answer 1

You have your equation already in vertex form.

Vertex form of a parabola equation is:
y = a(x - h)² + k

In this form, the vertex is the point (h, k). And the value of 'a' tells you whether the parabola opens upward (positive 'a') or downward (negative 'a'). If it opens upwards, the vertex is a minimum (bottom of the curve). If it is negative, the vertex is a maximum (top of the curve).

In your example you have:
y = 3(x + 1)² - 4

The only thing you have to do is to get the signs correct so it matches the vertex form:
y = a(x - h)² + k
y = 3(x - (-1))² + (-4)

Now read off the values of a, h, and k:
a = 3
h = -1
k = -4

The vertex is (-1, -4) and this point is the *minimum* point on the parabola (because 3 is positive and therefore it opens upward).

Answer 2

Take the first derivative of the equation with respect to x and set it equal to zero:

3x² + 6x - 1

6x + 6 = 0
6x = -6
x = -1

Now, compute y from that:

y = 3(-1 + 1)² - 4
y = 3(0) - 4
y = -4

You now have the vertex = (-1, -4).

Now, figure y for -2 and 0:

y = 3(-2 + 1)² - 4
y = 3(-1)² - 4
y = 3(1) - 4
y = -1

y = 3(0 + 1)² - 4
y = 3(1) - 4
y = 3 - 4
y = -1

Since y is greater at x = -2 and 0, it must be increasing on both sides. Therefore:

The Vertex is (-1, -4), and this point is the lowest point on the parabola.

Answer 3

I think that is 0, -4 and I know it is the minimum or lowest vlue because there is no negative in front