QuestionSketch the graph of $f(x)=3(x+1)^{2}-1$ , find the axis of symmetry, domain, and range. Use vertex and intercepts.

Studdy Solution

STEP 1

Assumptions1. The function is a quadratic function given by $f(x)=3(x+1)^{}-1$ . The vertex form of a quadratic function is $f(x)=a(x-h)^{}+k$ , where $(h,k)$ is the vertex of the parabola3. The axis of symmetry of a parabola in the form $f(x)=a(x-h)^{}+k$ is the line $x=h$
4. The domain of a quadratic function is all real numbers5. The range of a quadratic function in the form $f(x)=a(x-h)^{}+k$ is $y\geq k$ if $a>0$ and $y\leq k$ if $a<0$

STEP 2

First, we identify the vertex of the parabola. The vertex is given by the point $(h,k)$ , where $h$ is the value of $x$ that makes the expression inside the square zero, and $k$ is the constant term.
$Vertex = (-1, -1)$

STEP 3

Next, we find the $y$ -intercept of the parabola. The $y$ -intercept is the value of $f(x)$ when $x=0$ .
$f(0) =3(0+1)^{2}-1$

STEP 4

Calculate the $y$ -intercept.
$f(0) =3(0+1)^{2}-1 =3(1)^{2}-1 =3-1 =2$ So, the $y$ -intercept is $2$ .

STEP 5

The axis of symmetry of the parabola is the line $x=h$ . In this case, $h=-1$ , so the axis of symmetry is the line $x=-1$ .

STEP 6

The domain of the function is all real numbers, since there are no restrictions on the values that $x$ can take.

STEP 7

The range of the function is determined by the vertex and the direction of the parabola. Since the coefficient of $(x+1)^{2}$ is positive, the parabola opens upwards. Therefore, the range is $y\geq k$ , where $k$ is the $y$ -coordinate of the vertex.
$Range = y \geq -1$

STEP 8

Now, we can sketch the graph of the function. The vertex is at $(-1,-1)$ , the $y$ -intercept is at $(0,2)$ , and the axis of symmetry is the line $x=-1$ . The parabola opens upwards, and the domain is all real numbers, while the range is $y\geq -1$ .

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