Algebra 1 Worksheets | Quadratic Functions Worksheets - Free Printable

Problem: Graph the given equations of parabolas.

The task involves graphing four different equations of parabolas. Let's solve each one step by step.

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#### 1) \( y = -2x^2 + 12x - 16 \)

This is a quadratic equation in the standard form \( y = ax^2 + bx + c \), where:
- \( a = -2 \)
- \( b = 12 \)
- \( c = -16 \)

##### Step 1: Determine the vertex
The vertex of a parabola given by \( y = ax^2 + bx + c \) is at:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = -2 \) and \( b = 12 \):
\[
x = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3
\]
To find the \( y \)-coordinate of the vertex, substitute \( x = 3 \) back into the equation:
\[
y = -2(3)^2 + 12(3) - 16
\]
\[
y = -2(9) + 36 - 16
\]
\[
y = -18 + 36 - 16
\]
\[
y = 2
\]
So, the vertex is \( (3, 2) \).

##### Step 2: Determine the direction of the parabola
Since \( a = -2 \) (negative), the parabola opens downward.

##### Step 3: Find additional points
Choose some \( x \)-values around the vertex and calculate the corresponding \( y \)-values:
- For \( x = 0 \):
\[
y = -2(0)^2 + 12(0) - 16 = -16
\]
Point: \( (0, -16) \)

- For \( x = 1 \):
\[
y = -2(1)^2 + 12(1) - 16 = -2 + 12 - 16 = -6
\]
Point: \( (1, -6) \)

- For \( x = 2 \):
\[
y = -2(2)^2 + 12(2) - 16 = -2(4) + 24 - 16 = -8 + 24 - 16 = 0
\]
Point: \( (2, 0) \)

- For \( x = 4 \):
\[
y = -2(4)^2 + 12(4) - 16 = -2(16) + 48 - 16 = -32 + 48 - 16 = 0
\]
Point: \( (4, 0) \)

- For \( x = 5 \):
\[
y = -2(5)^2 + 12(5) - 16 = -2(25) + 60 - 16 = -50 + 60 - 16 = -6
\]
Point: \( (5, -6) \)

##### Step 4: Plot the points and sketch the parabola
Plot the vertex \( (3, 2) \) and the points \( (0, -16) \), \( (1, -6) \), \( (2, 0) \), \( (4, 0) \), and \( (5, -6) \). Draw a smooth curve through these points, opening downward.

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#### 2) \( y = -x^2 - 2x - 4 \)

This is another quadratic equation in the standard form \( y = ax^2 + bx + c \), where:
- \( a = -1 \)
- \( b = -2 \)
- \( c = -4 \)

##### Step 1: Determine the vertex
The vertex is at:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = -1 \) and \( b = -2 \):
\[
x = -\frac{-2}{2(-1)} = \frac{2}{-2} = -1
\]
To find the \( y \)-coordinate of the vertex, substitute \( x = -1 \) back into the equation:
\[
y = -(-1)^2 - 2(-1) - 4
\]
\[
y = -1 + 2 - 4
\]
\[
y = -3
\]
So, the vertex is \( (-1, -3) \).

##### Step 2: Determine the direction of the parabola
Since \( a = -1 \) (negative), the parabola opens downward.

##### Step 3: Find additional points
Choose some \( x \)-values around the vertex and calculate the corresponding \( y \)-values:
- For \( x = 0 \):
\[
y = -(0)^2 - 2(0) - 4 = -4
\]
Point: \( (0, -4) \)

- For \( x = -2 \):
\[
y = -(-2)^2 - 2(-2) - 4 = -4 + 4 - 4 = -4
\]
Point: \( (-2, -4) \)

- For \( x = 1 \):
\[
y = -(1)^2 - 2(1) - 4 = -1 - 2 - 4 = -7
\]
Point: \( (1, -7) \)

- For \( x = -3 \):
\[
y = -(-3)^2 - 2(-3) - 4 = -9 + 6 - 4 = -7
\]
Point: \( (-3, -7) \)

##### Step 4: Plot the points and sketch the parabola
Plot the vertex \( (-1, -3) \) and the points \( (0, -4) \), \( (-2, -4) \), \( (1, -7) \), and \( (-3, -7) \). Draw a smooth curve through these points, opening downward.

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#### 3) \( x = (y - 2)^2 - 3 \)

This is a horizontal parabola in the form \( x = a(y - k)^2 + h \), where:
- \( a = 1 \)
- \( h = -3 \)
- \( k = 2 \)

##### Step 1: Determine the vertex
The vertex of a horizontal parabola \( x = a(y - k)^2 + h \) is at \( (h, k) \). Here, the vertex is:
\[
(h, k) = (-3, 2)
\]

##### Step 2: Determine the direction of the parabola
Since \( a = 1 \) (positive), the parabola opens to the right.

##### Step 3: Find additional points
Choose some \( y \)-values around the vertex and calculate the corresponding \( x \)-values:
- For \( y = 0 \):
\[
x = (0 - 2)^2 - 3 = 4 - 3 = 1
\]
Point: \( (1, 0) \)

- For \( y = 1 \):
\[
x = (1 - 2)^2 - 3 = 1 - 3 = -2
\]
Point: \( (-2, 1) \)

- For \( y = 3 \):
\[
x = (3 - 2)^2 - 3 = 1 - 3 = -2
\]
Point: \( (-2, 3) \)

- For \( y = 4 \):
\[
x = (4 - 2)^2 - 3 = 4 - 3 = 1
\]
Point: \( (1, 4) \)

##### Step 4: Plot the points and sketch the parabola
Plot the vertex \( (-3, 2) \) and the points \( (1, 0) \), \( (-2, 1) \), \( (-2, 3) \), and \( (1, 4) \). Draw a smooth curve through these points, opening to the right.

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#### 4) \( x = -2(y + 3)^2 + 1 \)

This is another horizontal parabola in the form \( x = a(y - k)^2 + h \), where:
- \( a = -2 \)
- \( h = 1 \)
- \( k = -3 \)

##### Step 1: Determine the vertex
The vertex is at:
\[
(h, k) = (1, -3)
\]

##### Step 2: Determine the direction of the parabola
Since \( a = -2 \) (negative), the parabola opens to the left.

##### Step 3: Find additional points
Choose some \( y \)-values around the vertex and calculate the corresponding \( x \)-values:
- For \( y = -5 \):
\[
x = -2(-5 + 3)^2 + 1 = -2(-2)^2 + 1 = -2(4) + 1 = -8 + 1 = -7
\]
Point: \( (-7, -5) \)

- For \( y = -4 \):
\[
x = -2(-4 + 3)^2 + 1 = -2(-1)^2 + 1 = -2(1) + 1 = -2 + 1 = -1
\]
Point: \( (-1, -4) \)

- For \( y = -2 \):
\[
x = -2(-2 + 3)^2 + 1 = -2(1)^2 + 1 = -2(1) + 1 = -2 + 1 = -1
\]
Point: \( (-1, -2) \)

- For \( y = -1 \):
\[
x = -2(-1 + 3)^2 + 1 = -2(2)^2 + 1 = -2(4) + 1 = -8 + 1 = -7
\]
Point: \( (-7, -1) \)

##### Step 4: Plot the points and sketch the parabola
Plot the vertex \( (1, -3) \) and the points \( (-7, -5) \), \( (-1, -4) \), \( (-1, -2) \), and \( (-7, -1) \). Draw a smooth curve through these points, opening to the left.

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Final Answer:

The graphs for all four equations are sketched as described above. The final answer is:

\[
\boxed{\text{Graphs are completed as explained.}}
\]