Question

1. opens up or down, and passes through (11, 15), (7, 7), and (4, 22)

Explanation:

To determine the quadratic function \( y = ax^2 + bx + c \) that opens up or down and passes through the points \((11, 15)\), \((7, 7)\), and \((4, 22)\), we can set up a system of equations.

Step 1: Substitute the points into the quadratic equation

For the point \((11, 15)\):

$$ 15 = a(11)^2 + b(11) + c \implies 121a + 11b + c = 15 \quad (1) $$

For the point \((7, 7)\):

$$ 7 = a(7)^2 + b(7) + c \implies 49a + 7b + c = 7 \quad (2) $$

For the point \((4, 22)\):

$$ 22 = a(4)^2 + b(4) + c \implies 16a + 4b + c = 22 \quad (3) $$

Step 2: Subtract equations to eliminate \( c \)

Subtract equation \((2)\) from equation \((1)\):

$$ (121a + 11b + c) - (49a + 7b + c) = 15 - 7 $$

$$ 72a + 4b = 8 \implies 18a + b = 2 \quad (4) $$

Subtract equation \((3)\) from equation \((2)\):

$$ (49a + 7b + c) - (16a + 4b + c) = 7 - 22 $$

$$ 33a + 3b = -15 \implies 11a + b = -5 \quad (5) $$

Step 3: Solve the system of linear equations

Subtract equation \((4)\) from equation \((5)\):

$$ (11a + b) - (18a + b) = -5 - 2 $$

$$ -7a = -7 \implies a = 1 $$

Substitute \( a = 1 \) into equation \((4)\):

$$ 18(1) + b = 2 \implies 18 + b = 2 \implies b = -16 $$

Step 4: Find \( c \)

Substitute \( a = 1 \) and \( b = -16 \) into equation \((3)\):

$$ 16(1) + 4(-16) + c = 22 $$

$$ 16 - 64 + c = 22 \implies -48 + c = 22 \implies c = 70 $$

Step 5: Determine the direction the parabola opens

The quadratic function is \( y = x^2 - 16x + 70 \). The coefficient of \( x^2 \) is \( a = 1 \), which is positive. So, the parabola opens up.

Final Answer

The quadratic function is \( y = x^2 - 16x + 70 \) and it opens up.

Answer:

To determine the quadratic function \( y = ax^2 + bx + c \) that opens up or down and passes through the points \((11, 15)\), \((7, 7)\), and \((4, 22)\), we can set up a system of equations.

Step 1: Substitute the points into the quadratic equation

For the point \((11, 15)\):

$$ 15 = a(11)^2 + b(11) + c \implies 121a + 11b + c = 15 \quad (1) $$

For the point \((7, 7)\):

$$ 7 = a(7)^2 + b(7) + c \implies 49a + 7b + c = 7 \quad (2) $$

For the point \((4, 22)\):

$$ 22 = a(4)^2 + b(4) + c \implies 16a + 4b + c = 22 \quad (3) $$

Step 2: Subtract equations to eliminate \( c \)

Subtract equation \((2)\) from equation \((1)\):

$$ (121a + 11b + c) - (49a + 7b + c) = 15 - 7 $$

$$ 72a + 4b = 8 \implies 18a + b = 2 \quad (4) $$

Subtract equation \((3)\) from equation \((2)\):

$$ (49a + 7b + c) - (16a + 4b + c) = 7 - 22 $$

$$ 33a + 3b = -15 \implies 11a + b = -5 \quad (5) $$

Step 3: Solve the system of linear equations

Subtract equation \((4)\) from equation \((5)\):

$$ (11a + b) - (18a + b) = -5 - 2 $$

$$ -7a = -7 \implies a = 1 $$

Substitute \( a = 1 \) into equation \((4)\):

$$ 18(1) + b = 2 \implies 18 + b = 2 \implies b = -16 $$

Step 4: Find \( c \)

Substitute \( a = 1 \) and \( b = -16 \) into equation \((3)\):

$$ 16(1) + 4(-16) + c = 22 $$

$$ 16 - 64 + c = 22 \implies -48 + c = 22 \implies c = 70 $$

Step 5: Determine the direction the parabola opens

The quadratic function is \( y = x^2 - 16x + 70 \). The coefficient of \( x^2 \) is \( a = 1 \), which is positive. So, the parabola opens up.

Final Answer

The quadratic function is \( y = x^2 - 16x + 70 \) and it opens up.