Algebra 2 help?

Question

1) parabola has a line of symmetry x = -5. The minimum value of the quadratic function that it represents is -7.

Find a possible equation of this parabola and explain how you found it.

2)Complete the square for the following quadratic function:

y = -6x2 + 36x - 12

Then, graph the function. Point out the line of symmetry and state the maximum or minimum value

3) The area of a trapezoid with height z cm is 20 sq. cm. The lengths of the parallel sides are (2z + 3) cm and (6z – 1) cm. Find the height of the trapezoid. In your final answer, include all of the formulas and calculations necessary to solve for the height

Answer 1

2)

Completing the square Method :

The quadratic function  y = -6x² + 36x - 12.

Take out common factor.

y = -6(x² - 6x + 2)

To change the expression (x² - 6x) into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

 Here x coefficient = -6. So, (half the x coefficient)² = (-6/2)² = 9.

Add and subtract 9 to the expression.

y = - 6(x² - 6x + 2 + 9 - 9)

y = - 6(x² - 6x + 9 - 7)

y = - 6(x² - 2(3x) + 3²) + 42

Apply Perfect Square Trinomial : u² - 2uv + v² = (u - v)².

y = - 6(x - 3)² + 42

The solution is y = - 6(x - 3)² + 42.

 

Comments

Contd...

Graph

The quadratic function y = -6x² + 36x - 12

y = ax² + bx + c

a= -6, b = 36, c = -12

Axis of symmetry x = -b/2a = -36/2(-6)

Line of symmetry x = 3

The function vertex form y = 6(x - 3)² +42

compare it to vertex form of parabola y = a(x - h)² + k

Maximum value of the function (h, k) = (3, 42)

Choose random values for y and find the corresponding values for x.

y	y = 6(x - 3)2 +42	(x, y)
1	y = - 6(1 - 3)2+ 42	(1, 18)
-1	y = -6(-1 - 3)2+42	(-1, -54)
4	y = - 6(4 - 3)2+ 42	(4, 36)
5	y = - 6(5 - 3)2+ 42	(5, 18)
6	y = - 6(6 - 3)2+ 42	(6, -12)

1.Draw a coordinate plane.

2.Plot the coordinate points.

3.Then sketch the graph, connecting the points with a smooth curve.

Answer 2

3) Area of the trapezoid A = (a + b)h/2

Where a, b are parallel sides and h is height.

in this case  h = z

 a = 2z + 3 and b = 6z - 1

A = 20 square cm

Area of the trapezoid

20 = (2z + 3 + 6z - 1)z/2

40 = (8z + 2)z

40 = 8z² + 2z

Divide each side by 2.

20 = 4z² + z

Now solve the equation 4z² + z - 20 = 0

Compare it to

Roots are

The solutions for height.

Consider positive value z = 2.11 cm.