1. Quadratic Applications Special Topics: Fall 2015

2. Number Problems Find two consecutive integers whose product is 22 more than 8 times the larger. x = smaller integer x + 1 = larger integer x (x +1) = 8(x + 1) + 22 x 2 + x = 8x + 8 + 22 x 2 – 7x – 30 = 0 (x – 10)(x + 3) = 0 x – 10 = 0 x = 10 (smaller integer) x + 1 = 11 (larger integer) x + 3 = 0 x = -3 (smaller integer) x + 1 = -2 (larger integer)

3. Number Problems Find an integer where three times the square is 20 more than 11 times the number. x = integer 3x 2 = 11x + 20 3x 2 – 11x – 20 = 0 (set equation to 0) Factor using slip and slide 3x 2 – 11x – 20 x 2 – 11x – 60 (slip a and multiply c) (x – 15) (x + 4) Factor (x – 15/3) (x + 4/3) Divide by a (x – 5) (x + 4/3) Simplify (x – 5) (3x + 4) Slide the denominator x – 5= 0 x = 5 3x + 4 = 0 3x = -4 x = -4/3 Check answer meets solution criteria x = -4/3 is not an integer x = 5

4. Area Problems A rectangle is 8 inches longer than it is wide. If the area of the rectangle is 105 square inches find the dimensions of the rectangle. x x + 8 x(x + 8) = 105 x = width of rectangle x + 8 = length of rectangle x 2 + 8x = 105 x 2 + 8x – 105 = 0 (x + 15)(x – 7) = 0 x + 15 = 0 -> x = -15 no negative x – 7 = 0 -> x = 7 x = 7 (width) x + 7 = 15 (length)

5. Area Problems A rectangle is 4 inches longer than it is wide. If the length is increased by 5 and the width is decreased by 2 then the area is 26 square inches. Find the dimensions of the original rectangle. x x + 4 x + 9 x – 2 (x – 2)(x + 9) = 26 x = width of original rectangle x + 4 = length of original rectangle x 2 + 9x – 2x – 18 = 26 x 2 + 7x – 44 = 0 (x + 11)(x – 4) = 0 x + 11 = 0 -> x = -11 no negative x – 4 = 0 -> x = 4 x = 4 width of original x + 4 = 8 length of original

6. Area Problems A photo is 5 inches by 7 inches. It is put into a frame that is x inches on each side. The total area of the photo with frame is 99 square inches. How wide is the frame? 5 7 x + 7 + x -> 2x + 7 x + 5 + x 2x + 5 (2x + 5) (2x + 7) = 99 x = width of frame 4x 2 + 14x + 10x + 35 = 99 4x 2 + 24x – 64 = 0 4(x 2 + 6x – 16) = 0 4(x + 8)(x – 2)= 0 x + 8 = 0 -> x = -8 no negatives x = 2 width of frame -x- |x||x| |x||x| x – 2 = 0 -> x = 2

7. Projectile Motion A model rocket is launched with an initial velocity of 178 ft/sec. Its height can be modeled as h(t) = -16t 2 + 178t What is the maximum height of the rocket? When will it reach the maximum height? When will it hit the ground? Use Y= and enter equation: -16x 2 + 178x Use 2 nd -Trace and find the maximum. Left of vertex for Left Bound. Right of vertex for Right Bound. Enter for Guess. Y-value = 495.06 X-value of maximum is 5.56 seconds Use 2 nd -Trace and find the zero. Left of zero (positive) for Left Bound. Right of zero (negative) for Right Bound. Enter for Guess. X = 11.13 s

8. Projectile Motion A water balloon is thrown off of a building and can be modeled h(t) = -16t 2 + 32t + 105 When will the water balloon hit the ground? What is the height after 3 seconds? How tall is the building? Use Y= and enter equation: -16x 2 + 32x + 105 Use 2 nd -Trace and find zero. X = 3.75 Use 2 nd -Trace and find value. X = 3 -> Y = 57 Look at c in the equation. This is initial height or 105 feet.

9. Quadratics Review Axis of Symmetry:______________________________________ Vertex:___________________________________________________ Domain:__________________________________________________ Range:____________________________________________________ Zeros:_____________________________________________________ x = -2 (-2, 1) All Real #s y ≤ 1 {-3, -1}