y = - 4x² - 16x - 7

y = - 4(x² + 4x) - 7

y = - 4(x² + 4x + 4 - 4) - 7

y = - 4(x² + 4x + 4) + 16 - 7

y = - 4(x + 2)² + 9

The vertex is (- 2, 9). The axis of symmetry is the line with the equation x = - 2.

Plot the vertex and draw the axis of symmetry.

Find another point.

When x = 0, y = - 4(0 + 2)² + 9 = - 16 + 9 = - 7. The point is (0, - 7).

The reflection of (0, - 7) is (- 4, - 7). Plot it.

Connect the points with a smooth curve.

y = x² + 4x + 1

y = (x² + 4x) + 1

y = (x² + 4x + 4 - 4) + 1

y = (x² + 4x + 4) - 4 + 1

y = (x + 2)² - 3

The vertex is (- 2, - 3). The axis of symmetry is the line with the equation x = - 2.

Plot the vertex and draw the axis of symmetry.

Find another point.

When x = 0, y = (0 + 2)² - 3 = 4 - 3 = 1 The point is (0, 1).

The reflection of (0, 1) is (- 4, 1). Plot it.

Connect the points with a smooth curve.

y = (2350)x² - (235)x + 12

y = (2350)(x² - 10x) + 12

y = (2350)(x² - 10x + 25 - 25) + 12

y = (2350)(x² -10x + 25) - (23×2550) + 12

y = (2350)(x² - 10x + 25) - (232) + 12

y = (2350)(x - 5)² + (12)

The vertex is (5, 12). The axis of symmetry is the line with the equation x = 5.

Plot the vertex and draw the axis of symmetry.

Find another point.

When x = 0, y = (2350)(0 - 5)² + (12) = (232) + (12) = 12 The point is (0, 12).

The reflection of (0, 12) is (10, 12). Plot it.

Connect the points with a smooth curve.

x² + y² = 2

The general form of the equation of the circle is (x - h)² + (y - k)² = r², where (x, y) are the coordinates of a point on the circle, (h, k) is the centre of the circle, and r is the radius of the circle.

Compare the given equation with the general form of the circle equation.

(h, k) = (0, 0)

r² = 2

r = 1.4

From the given graphs, we can observe that the circle in Graph 1 has centre as origin and radius as 1.4.

Therefore, Graph 1 represents the given equation of the circle.

x² + y² = 7

The general form of the equation of the circle is (x - h)² + (y - k)² = r², where (x, y) are the coordinates of a point on the circle, (h, k) is the centre of the circle, and r is the radius of the circle.

Compare the given equation with the general form of the circle equation.

(h, k) = (0, 0)

r² = 7

r = 2.7

From the given graphs, we can observe that the circle in Graph 2 has centre as origin and radius as 2.7.

Therefore, Graph 2 represents the given equation of the circle.

x² + y² = 9

The general form of the equation of the circle is (x - h)² + (y - k)² = r², where (x, y) are the coordinates of a point on the circle, (h, k) is the centre of the circle, and r is the radius of the circle.

Compare the given equation with the general form of the circle equation.

(h, k) = (0, 0)

r² = 9

r = 3

From the given graphs, we can observe that the circle in Graph 1 has centre as origin and radius as 3.

Therefore, Graph 1 represents the given equation of the circle.

x² + y² = 6

The general form of the equation of the circle is (x - h)² + (y - k)² = r², where (x, y) are the coordinates of a point on the circle, (h, k) is the centre of the circle, and r is the radius of the circle.

Compare the given equation with the general form of the circle equation.

(h, k) = (0, 0)

r² = 6

r = 2.4

From the given graphs, we can observe that the circle in Graph 2 has centre as origin and radius as 2.4.

Therefore, Graph 2 represents the given equation of the circle.

x210 + y25 = 1

The standard form of the equation of ellipse is (x - h)2a2 + (y - k)2b 2 = 1.

Compare the given equation with the standard form of the equation of ellipse.

The center of the ellipse is at (0, 0).

a² = 10

a = 10

b² = 5

b = 5

Length of the major axis = 2a
= 210 units

Length of the minor axis = 2b
= 25 units

c² = a² - b²

c² = 10 - 5 = 5

c = 5

So, the foci of the ellipse are (5, 0) and (- 5, 0).

From the given graphs, we can observe that the ellipse in Graph 2 have center at the origin, foci as (5, 0) and (- 5, 0), length of the major axis as 210 units minor axis as 25 units, x-intercepts as (± 10, 0), and y-intercepts as (± 5, 0).

Therefore, Graph 2 represents the given equation of the ellipse.

4x2 + 8y2 = 32

x28 + y24 = 1

The standard form of the equation of ellipse is (x - h)2a2 + (y - k)2b 2 = 1.

Compare the given equation with the standard form of the equation of ellipse.

The center of the ellipse is at (0, 0).

a² = 8

a = 22

b² = 4

b = 2

Length of the major axis = 2a
= 2(22) = 42 units

Length of the minor axis = 2b
= 2(2) = 4 units

c² = a² - b²

c² = 8 - 4 = 4

c = 2

So, the foci of the ellipse are (2, 0) and (- 2, 0).

From the given graphs, we can observe that the ellipse in Graph 4 have center at the origin, foci as (2, 0) and (- 2, 0), length of the major axis as 42 units and minor axis as 4 units, x-intercepts as (± 22, 0), and y-intercepts as (± 2, 0).

Therefore, Graph 4 represents the given equation of the ellipse.

x2 + 36y2 = 36

x236 + y21 = 1

The standard form of the equation of ellipse is (x - h)2a2 + (y - k)2b 2 = 1.

Compare the given equation with the standard form of the equation of ellipse.

The center of the ellipse is at (0, 0).

a² = 36

a = 6

b² = 1

b = 1

Length of the major axis = 2a
= 2(6) = 12 units

Length of the minor axis = 2b
= 2(1) = 2 units

c² = a² - b²

c² = 36 - 1 = 35

c = 35

So, the foci of the ellipse are (35, 0) and (- 35, 0).

From the given graphs, we can observe that the ellipse in Graph 3 have center at the origin, foci as (35, 0) and (- 35, 0), length of the major axis as 12 units and minor axis as 2 units, x-intercepts as (± 6, 0), and y-intercepts as (± 1, 0).

Therefore, Graph 3 represents the given equation of the ellipse.

x2 + 2y2 = 2

x22 + y21 = 1

The standard form of the equation of ellipse is (x - h)2a2 + (y - k)2b 2 = 1.

Compare the given equation with the standard form of the equation of ellipse.

The center of the ellipse is at (0, 0).

a² = 2

a = 2

b² = 1

b = 1

Length of the major axis = 2a
= 22 units

Length of the minor axis = 2b
= 2(1) = 2 units

c² = a² - b²

c² = 2 - 1 = 1

c = 1

So, the foci of the ellipse are (1, 0) and (- 1, 0).

From the given graphs, we can observe that the ellipse in Graph 2 have center at the origin, foci as (1, 0) and (- 1, 0), length of the major axis as 2rout(2) units and minor axis as 2 units, x-intercepts as (± 2, 0), and y-intercepts as (± 1, 0).

Therefore, Graph 2 represents the given equation of the ellipse.

The standard form of the equation x2a2 - y2b2 = 1.

Compare x² - 2 = 2y² to the standard form of the hyperbola equation.

The center of this hyperbola is at the origin. According to the equation, a² = 2, so a = ±2 and b² = 1, so b = ±1.
The coordinates of the vertices are (2, 0) and (- 2, 0).

So, the hyperbola passes through the points (2, 0) and (- 2, 0).

The points (0, - 1) and (0, 1) and the points (2, 0) and (- 2, 0) can be used to form a rectangle whose diagonals are the asymptotes of the hyperbola.

Now we can sketch the hyperbola. The graph goes through the points (2, 0) and (- 2, 0) and get closer and closer to the asymptotes as we get farther and farther from the center of the figure.

Therefore, Graph 2 represents the equation x² - 2 = 2y².

The standard form of the equation x2a2 - y2b2 = 1.

Compare x29 - y216 = 1 to the standard form of the hyperbola equation.

The center of this hyperbola is at the origin. According to the equation, a² = 9, so a = ±3 and b² = 16, so b = ±4.
The coordinates of the vertices are (3, 0) and (- 3, 0).

So, the hyperbola passes through the points (3, 0) and (- 3, 0).

The points (0, - 4) and (0, 4) and the points (3, 0) and (- 3, 0) can be used to form a rectangle whose diagonals are the asymptotes of the hyperbola.

Now we can sketch the hyperbola. The graph goes through the points (3, 0) and (- 3, 0) and get closer and closer to the asymptotes as we get farther and farther from the center of the figure.

Therefore, Graph 3 represents the equation x29 - y216 = 1.

The standard form of the equation x2a2 - y2b2 = 1.

Compare 3x2 - 9y2 = 27 with the standard form of the hyperbola equation.

The center of this hyperbola is at the origin. According to the equation, a² = 9, so a = ±3 and b² = 3, so b = ±3.
The coordinates of the vertices are (3, 0) and (- 3, 0).

So, the hyperbola passes through the points (3, 0) and (- 3, 0).

The points (0, - 3) and (0, 3) and the points (3, 0) and (- 3, 0) can be used to form a rectangle whose diagonals are the asymptotes of the hyperbola.

Now we can sketch the hyperbola. The graph goes through the points (3, 0) and (- 3, 0) and get closer and closer to the asymptotes as we get farther and farther from the center of the figure.

Therefore, Graph 1 represents the equation 3x2 - 9y2 = 27.

The standard form of the equation x2a2 - y2b2 = 1.

Compare 25x2 - 9y2 = 225 with the standard form of the hyperbola equation.

The center of this hyperbola is at the origin. According to the equation, a² = 9, so a = ±3 and b² = 25, so b = ±5.
The coordinates of the vertices are (3, 0) and (- 3, 0).

So, the hyperbola passes through the points (3, 0) and (- 3, 0).

The points (0, - 5) and (0, 5) and the points (3, 0) and (- 3, 0) can be used to form a rectangle whose diagonals are the asymptotes of the hyperbola.

Now we can sketch the hyperbola. The graph goes through the points (3, 0) and (- 3, 0) and get closer and closer to the asymptotes as we get farther and farther from the center of the figure.

Therefore, Graph 2 represents the equation 25x2 - 9y2 = 225.