Question2. $3(x+1)^{2}-3=0$ . .xis of symmetry: Vertex: Solution(s):

Studdy Solution

STEP 1

What is this asking? We're asked to solve a quadratic equation, find its axis of symmetry, and the vertex! Watch out! Don't forget to consider both positive and negative square roots when solving!

STEP 2

1. Rewrite the equation
2. Solve for $x$
3. Find the axis of symmetry
4. Find the vertex

STEP 3

Let's isolate the $(x+1)^2$ term.
Add 3 to both sides of the equation: $3(x+1)^2 - 3 + 3 = 0 + 3$ $3(x+1)^2 = 3$ Now, divide both sides by 3: $\frac{3(x+1)^2}{3} = \frac{3}{3}$ $(x+1)^2 = 1$

STEP 4

Remember, when we take the square root, we get two possible solutions: a positive and a negative one! $\sqrt{(x+1)^2} = \pm\sqrt{1}$ $x+1 = \pm 1$

STEP 5

First, let's consider the positive root: $x+1 = 1$ Subtract 1 from both sides: $x+1-1 = 1-1$ $x = 0$ Now, let's consider the negative root: $x+1 = -1$ Subtract 1 from both sides: $x+1-1 = -1-1$ $x = -2$ So, our two solutions are $x=0$ and $x=-2$ !

STEP 6

The axis of symmetry is right in the middle of our two solutions.
To find the middle, we take the average of the two solutions: $x = \frac{0 + (-2)}{2}$ $x = \frac{-2}{2}$ $x = -1$ So, the axis of symmetry is $x = -1$ !

STEP 7

The x-coordinate of the vertex is the same as the axis of symmetry, which we just found to be $x=-1$ .

STEP 8

To find the y-coordinate, plug $x=-1$ back into our original equation: $y = 3((-1)+1)^2 - 3$ $y = 3(0)^2 - 3$ $y = 3 \cdot 0 - 3$ $y = 0 - 3$ $y = -3$ So, the vertex is at $(-1, -3)$ !

STEP 9

The solutions to the equation $3(x+1)^2 - 3 = 0$ are $x=0$ and $x=-2$ .
The axis of symmetry is $x=-1$ , and the vertex is at $(-1, -3)$ .

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