Math / Algebra

QuestionGraph the following polynomials find at least 5 points Part two: Write each of the equations in standard form and place next to the vertex form. Please sho the AOS of the parabola 1) $y=-(x-2)^{2}-2$ S.F $\qquad$

Studdy Solution

STEP 1

1. We are given a quadratic equation in vertex form: $y = -(x-2)^2 - 2$ .
2. We need to graph this polynomial and find at least 5 points on the graph.
3. We need to convert the given equation to standard form.
4. We need to find the axis of symmetry (AOS) of the parabola.

STEP 2

1. Identify key features of the vertex form of the quadratic.
2. Convert the vertex form to standard form.
3. Determine the axis of symmetry.
4. Calculate and plot at least 5 points on the graph.
5. Sketch the graph of the polynomial.

STEP 3

The given equation $y = -(x-2)^2 - 2$ is in vertex form $y = a(x-h)^2 + k$ .
The vertex of the parabola is $(h, k) = (2, -2)$ .
The parabola opens downwards because $a = -1$ is negative.

STEP 4

Expand the vertex form to convert it to standard form.
$y = -(x-2)^2 - 2$
$y = -(x^2 - 4x + 4) - 2$
$y = -x^2 + 4x - 4 - 2$
$y = -x^2 + 4x - 6$
The standard form is $y = -x^2 + 4x - 6$ .

STEP 5

The axis of symmetry (AOS) for a parabola in vertex form $y = a(x-h)^2 + k$ is $x = h$ .
For the given equation, the AOS is $x = 2$ .

STEP 6

Calculate and plot at least 5 points on the graph, including the vertex.
1. Vertex: $(2, -2)$
2. Choose $x = 1$ : $y = -(1-2)^2 - 2 = -1 - 2 = -3$ $\Rightarrow (1, -3)$
3. Choose $x = 3$ : $y = -(3-2)^2 - 2 = -1 - 2 = -3$ $\Rightarrow (3, -3)$
4. Choose $x = 0$ : $y = -(0-2)^2 - 2 = -4 - 2 = -6$ $\Rightarrow (0, -6)$
5. Choose $x = 4$ : $y = -(4-2)^2 - 2 = -4 - 2 = -6$ $\Rightarrow (4, -6)$

STEP 7

Sketch the graph using the points:
- Plot the vertex $(2, -2)$ . - Plot the points $(1, -3)$ , $(3, -3)$ , $(0, -6)$ , and $(4, -6)$ . - Draw a smooth curve through these points to form the parabola.
The equation in standard form is $y = -x^2 + 4x - 6$ .
The axis of symmetry is $x = 2$ .

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QuestionGraph the following polynomials find at least 5 points Part two: Write each of the equations in standard form and place next to the vertex form. Please sho the AOS of the parabola 1) $y=-(x-2)^{2}-2$ S.F $\qquad$

Studdy Solution

STEP 1

1. We are given a quadratic equation in vertex form: $y = -(x-2)^2 - 2$ .
2. We need to graph this polynomial and find at least 5 points on the graph.
3. We need to convert the given equation to standard form.
4. We need to find the axis of symmetry (AOS) of the parabola.

STEP 2

1. Identify key features of the vertex form of the quadratic.
2. Convert the vertex form to standard form.
3. Determine the axis of symmetry.
4. Calculate and plot at least 5 points on the graph.
5. Sketch the graph of the polynomial.

STEP 3

The given equation $y = -(x-2)^2 - 2$ is in vertex form $y = a(x-h)^2 + k$ .
The vertex of the parabola is $(h, k) = (2, -2)$ .
The parabola opens downwards because $a = -1$ is negative.

STEP 4

Expand the vertex form to convert it to standard form.
$y = -(x-2)^2 - 2$
$y = -(x^2 - 4x + 4) - 2$
$y = -x^2 + 4x - 4 - 2$
$y = -x^2 + 4x - 6$
The standard form is $y = -x^2 + 4x - 6$ .

STEP 5

The axis of symmetry (AOS) for a parabola in vertex form $y = a(x-h)^2 + k$ is $x = h$ .
For the given equation, the AOS is $x = 2$ .

STEP 6

STEP 7

Sketch the graph using the points:
- Plot the vertex $(2, -2)$ . - Plot the points $(1, -3)$ , $(3, -3)$ , $(0, -6)$ , and $(4, -6)$ . - Draw a smooth curve through these points to form the parabola.
The equation in standard form is $y = -x^2 + 4x - 6$ .
The axis of symmetry is $x = 2$ .

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QuestionGraph the following polynomials find at least 5 points Part two: Write each of the equations in standard form and place next to the vertex form. Please sho the AOS of the parabola 1) y=−(x−2)2−2y=-(x-2)^{2}-2y=−(x−2)2−2 S.F \qquad

Studdy Solution

STEP 1

STEP 2

1. Identify key features of the vertex form of the quadratic.2. Convert the vertex form to standard form.3. Determine the axis of symmetry.4. Calculate and plot at least 5 points on the graph.5. Sketch the graph of the polynomial.

STEP 3

The given equation y=−(x−2)2−2 y = -(x-2)^2 - 2 y=−(x−2)2−2 is in vertex form y=a(x−h)2+k y = a(x-h)^2 + k y=a(x−h)2+k.The vertex of the parabola is (h,k)=(2,−2) (h, k) = (2, -2) (h,k)=(2,−2).The parabola opens downwards because a=−1 a = -1 a=−1 is negative.

STEP 4

STEP 5

The axis of symmetry (AOS) for a parabola in vertex form y=a(x−h)2+k y = a(x-h)^2 + k y=a(x−h)2+k is x=h x = h x=h.For the given equation, the AOS is x=2 x = 2 x=2.

STEP 6

STEP 7

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QuestionGraph the following polynomials find at least 5 points Part two: Write each of the equations in standard form and place next to the vertex form. Please sho the AOS of the parabola 1) $y=-(x-2)^{2}-2$ S.F $\qquad$

1. Identify key features of the vertex form of the quadratic.
2. Convert the vertex form to standard form.
3. Determine the axis of symmetry.
4. Calculate and plot at least 5 points on the graph.
5. Sketch the graph of the polynomial.

The given equation $y = -(x-2)^2 - 2$ is in vertex form $y = a(x-h)^2 + k$ .
The vertex of the parabola is $(h, k) = (2, -2)$ .
The parabola opens downwards because $a = -1$ is negative.

The axis of symmetry (AOS) for a parabola in vertex form $y = a(x-h)^2 + k$ is $x = h$ .
For the given equation, the AOS is $x = 2$ .