Math / Algebra

QuestionQuadratic and Exponential Functions Graphing a parabola of the form $y=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ : Integer coefficier aph the parabola. $y=3 x^{2}-30 x+69$
Plot five points on the parabola: the vertex, two points to the le button.

Studdy Solution

STEP 1

1. The given quadratic function is $y = 3x^2 - 30x + 69$ .
2. The vertex of the parabola can be found using the vertex formula.
3. The parabola is symmetric about its vertex.
4. The axis of symmetry can be used to find points equidistant from the vertex.
5. Integer coefficients suggest integer calculations for simplicity.

STEP 2

1. Find the vertex of the parabola.
2. Determine the axis of symmetry.
3. Calculate two points to the left of the vertex.
4. Calculate two points to the right of the vertex.
5. Plot the points and sketch the parabola.

STEP 3

Find the vertex of the parabola using the vertex formula. The x-coordinate of the vertex is given by:
$x = -\frac{b}{2a}$
For the equation $y = 3x^2 - 30x + 69$ , $a = 3$ and $b = -30$ . Substitute these values into the formula:
$x = -\frac{-30}{2 \times 3} = \frac{30}{6} = 5$

STEP 4

Substitute $x = 5$ back into the original equation to find the y-coordinate of the vertex:
$y = 3(5)^2 - 30(5) + 69$ $y = 3(25) - 150 + 69$ $y = 75 - 150 + 69$ $y = -6$
So, the vertex is $(5, -6)$ .

STEP 5

The axis of symmetry is the vertical line that passes through the vertex. For this parabola, it is:
$x = 5$

STEP 6

Calculate two points to the left of the vertex. Choose $x = 3$ and $x = 4$ .
For $x = 3$ :
$y = 3(3)^2 - 30(3) + 69$ $y = 3(9) - 90 + 69$ $y = 27 - 90 + 69$ $y = 6$
For $x = 4$ :
$y = 3(4)^2 - 30(4) + 69$ $y = 3(16) - 120 + 69$ $y = 48 - 120 + 69$ $y = -3$
So, the points are $(3, 6)$ and $(4, -3)$ .

STEP 7

Calculate two points to the right of the vertex. Choose $x = 6$ and $x = 7$ .
For $x = 6$ :
$y = 3(6)^2 - 30(6) + 69$ $y = 3(36) - 180 + 69$ $y = 108 - 180 + 69$ $y = -3$
For $x = 7$ :
$y = 3(7)^2 - 30(7) + 69$ $y = 3(49) - 210 + 69$ $y = 147 - 210 + 69$ $y = 6$
So, the points are $(6, -3)$ and $(7, 6)$ .

STEP 8

Plot the points $(3, 6)$ , $(4, -3)$ , $(5, -6)$ , $(6, -3)$ , and $(7, 6)$ on a graph. Draw a smooth curve through these points to sketch the parabola.
The parabola is now graphed with the vertex and four additional points plotted.

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QuestionQuadratic and Exponential Functions Graphing a parabola of the form $y=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ : Integer coefficier aph the parabola. $y=3 x^{2}-30 x+69$
Plot five points on the parabola: the vertex, two points to the le button.

Studdy Solution

STEP 1

STEP 2

1. Find the vertex of the parabola.
2. Determine the axis of symmetry.
3. Calculate two points to the left of the vertex.
4. Calculate two points to the right of the vertex.
5. Plot the points and sketch the parabola.

STEP 3

Find the vertex of the parabola using the vertex formula. The x-coordinate of the vertex is given by:
$x = -\frac{b}{2a}$
For the equation $y = 3x^2 - 30x + 69$ , $a = 3$ and $b = -30$ . Substitute these values into the formula:
$x = -\frac{-30}{2 \times 3} = \frac{30}{6} = 5$

STEP 4

Substitute $x = 5$ back into the original equation to find the y-coordinate of the vertex:
$y = 3(5)^2 - 30(5) + 69$ $y = 3(25) - 150 + 69$ $y = 75 - 150 + 69$ $y = -6$
So, the vertex is $(5, -6)$ .

STEP 5

The axis of symmetry is the vertical line that passes through the vertex. For this parabola, it is:
$x = 5$

STEP 6

Calculate two points to the left of the vertex. Choose $x = 3$ and $x = 4$ .
For $x = 3$ :
$y = 3(3)^2 - 30(3) + 69$ $y = 3(9) - 90 + 69$ $y = 27 - 90 + 69$ $y = 6$
For $x = 4$ :
$y = 3(4)^2 - 30(4) + 69$ $y = 3(16) - 120 + 69$ $y = 48 - 120 + 69$ $y = -3$
So, the points are $(3, 6)$ and $(4, -3)$ .

STEP 7

STEP 8

Plot the points $(3, 6)$ , $(4, -3)$ , $(5, -6)$ , $(6, -3)$ , and $(7, 6)$ on a graph. Draw a smooth curve through these points to sketch the parabola.
The parabola is now graphed with the vertex and four additional points plotted.

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QuestionQuadratic and Exponential Functions Graphing a parabola of the form y=ax2+bx+cy=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}y=ax2+bx+c : Integer coefficier aph the parabola. y=3x2−30x+69y=3 x^{2}-30 x+69y=3x2−30x+69Plot five points on the parabola: the vertex, two points to the le button.

Studdy Solution

STEP 1

STEP 2

1. Find the vertex of the parabola.2. Determine the axis of symmetry.3. Calculate two points to the left of the vertex.4. Calculate two points to the right of the vertex.5. Plot the points and sketch the parabola.

STEP 3

STEP 4

STEP 5

The axis of symmetry is the vertical line that passes through the vertex. For this parabola, it is:x=5 x = 5 x=5

STEP 6

STEP 7

STEP 8

Plot the points (3,6) (3, 6) (3,6), (4,−3) (4, -3) (4,−3), (5,−6) (5, -6) (5,−6), (6,−3) (6, -3) (6,−3), and (7,6) (7, 6) (7,6) on a graph. Draw a smooth curve through these points to sketch the parabola. The parabola is now graphed with the vertex and four additional points plotted.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionQuadratic and Exponential Functions Graphing a parabola of the form $y=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ : Integer coefficier aph the parabola. $y=3 x^{2}-30 x+69$
Plot five points on the parabola: the vertex, two points to the le button.

1. Find the vertex of the parabola.
2. Determine the axis of symmetry.
3. Calculate two points to the left of the vertex.
4. Calculate two points to the right of the vertex.
5. Plot the points and sketch the parabola.

The axis of symmetry is the vertical line that passes through the vertex. For this parabola, it is:
$x = 5$

Plot the points $(3, 6)$ , $(4, -3)$ , $(5, -6)$ , $(6, -3)$ , and $(7, 6)$ on a graph. Draw a smooth curve through these points to sketch the parabola.
The parabola is now graphed with the vertex and four additional points plotted.