8-2 Quadratic Functions (Part # 1)
- Example 1
- Example 2 - Vertex Formula
- Example 3 + up/down test
- Example 4 Graph f(x) = x^2 -2x - 8
- Example 5: Graph y = x^2 + 2x + 3
- Example 6: Graph y = 2x^2 - 8x
- Example 7: h = 16t^2 + 72t + 520
Example 1
- A) The vertex is (4,3)
- B) The vertex is (-3,-3)
Example 2 - Vertex Formula
a) y = 2x^2+4x
-
-
we get the a, b, c in the formula y = ax^2 + bx + c
-
a = 2
-
b = 4
-
c = 0
-
-
then we plug in a, b into the formula
-
x = -b / (2a)
-
x = -4 / (2*2) = -4/4 = -1
-
-
-
so the axis of symmetry is x = -1
-
to find the vertex
-
so the known vertex data is (-1, y)
-
to find y for the vertex we need to plug -1 into the formula y = 2x^2+4x
-
y = 2*-1^2+4*-1 = -6
-
so the vertex is (-1, -6)
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
B) y = -x^2 + 4x - 5
-
-
we get the a, b, c in the formula y = ax^2 + bx + c
-
a = -1
-
b = 4
-
c = -5
-
-
then we plug in a, b into the formula
-
x = -b / (2a)
-
x = -4 / (2*-1) = -4/-2 = 2
-
-
-
so the axis of symmetry is x = 2
-
to find the vertex
-
so the known vertex data is (2, y)
-
to find y for the vertex we need to plug -1 into the formula y = 2x^2+4x
-
y = -2^2 + 4 * 2 - 5 = -1
-
so the vertex is (2, -1)
Example 3 + up/down test
up/down test
- opens upwards because a is Positive
- opens downward because a is Negative
Example 3
- a) y= x^2 + 3x + 4
- opens upwards because a is Positive
- opens upwards because a is Positive
- b) y = -3x^2 + 5x
- opens downward because a is Negative
- c) y = 2x - x^2 + 6
- opens downward because a is Negative
- opens downward because a is Negative
Example 4 Graph f(x) = x^2 -2x - 8
Steps to Graph ax^2 + bx + c
- Find the vertex and the axis of symmetry. Sketch these in.
- Find the x-intercept by plugging in 0 for y.
- Find the y-intercept by plugging in 0 for x.
- Reflect your points across the axis of symmetry and connect your dots with a smooth U-shaped (not V-shaped) curve.
Graph f(x) = x^2 -2x - 8
for now, I'm just gonna type my work and figure out what to do next
- find the line of symmetry -
- a = 1, b = -2, c = -8
- use this to find the vertex
- x = (b/2a)
- x = -(-2) / 2(1) = 1
- since we know that the along the x axis at 1 will be the vertex we replace x with 1 in the original formula
- x=1
- y = x^2 -2x - 8
- y = 1^2 + -2 * 1 - 8 = 1- 2 - 8 = -9
- y = -9
- x=1
- the vertex is (1, -9)
- since the vertex is -1,-9 we know that x=1 is the axis of symmetry
- finding the y-intercept is the easiest to start with because we just replace x with 0
- x = 0 | y = x^2 -2x - 8
- y = 0 - 8 = -8
- y-intercept = (0,-8)
- so so To find the x-intercepts, you can set y equal to zero and solve for x:
-
y = 0 | x = (-b ± sqrt(b^2 - 4ac)) / 2a
-
x = (-(-2) ± sqrt((-2)^2 - 4(1)(-8))) / 2(1)
x = (2 ± sqrt(4 + 32)) / 2
x = (2 ± sqrt(36)) / 2
x = (2 ± 6) / 2
x = 8 / 2 or x = -4/ 2
x = 4 or x = -2
sooooo (-2,0) & (4,0)
-
- so since we know 3 y axis points on the graph and the axis of symmetry we can get another point without doing much work
- symmetry line = x = 1,
- calc'd x-intercept 0,-8
- the symmetry line is 1 and the known point is 0 since 1-0 = 1 we can add that to the x coordinate of y and keep the same y coordinate to get the mirrored point making another point on the graph (2,-8)
- since we need one more point for the graph we can choose say x=3, | x^2 -2x - 8
- y = 3^2 -3*2 - 8 = -5
- soooo the new point is (3,-5) if we mirror that along 1,-9 we get (-1, -5 ) because 3 is 2 more than 1, and 2 less than 1 is -1. we also keep the same y coordinate
- y = 3^2 -3*2 - 8 = -5
- so all points are:
- (1, -9)
- (0,-8)
- (2, -8)
- (3,-5)
- (-1,-5)
Example 5: Graph y = x^2 + 2x + 3
Find the vertex and the axis of symmetry. Sketch these in.
• Find the x-intercept by plugging in 0 for y.
• Find the y-intercept by plugging in 0 for x.
• Reflect your points across the axis of symmetry and connect your dots with a smooth
U-shaped (not V-shaped) curve.
fix the fallowing
- a = 1, b = 2, c = 3
- x^2 + 2x + 3
- find the line of symmetry -
- x = (b/2a)
- x = -(2) / 2(1) =1
- use this to find the vertex
- since we know that the along the x axis at -1 will be the vertex we replace x with 1 in the original formula
- x=1
- y = x^2 -2x - 8
- y = 1^2 + -2 * 1 - 8 = 1- 2 - 8 = -9
- y = -9
- x=1
- the vertex is (1, -9)
- since the vertex is -1,-9 we know that x=1 is the axis of symmetry
- finding the y-intercept is the easiest to start with because we just replace x with 0
- x = 0 | y = x^2 -2x - 8
- y = 0 - 8 = -8
- y-intercept = (0,-8)
- so so To find the x-intercepts, you can set y equal to zero and solve for x:
-
y = 0 | x = (-b ± sqrt(b^2 - 4ac)) / 2a
-
x = (-(-2) ± sqrt((-2)^2 - 4(1)(-8))) / 2(1)
x = (2 ± sqrt(4 + 32)) / 2
x = (2 ± sqrt(36)) / 2
x = (2 ± 6) / 2
x = 8 / 2 or x = -4/ 2
x = 4 or x = -2
sooooo (-2,0) & (4,0)
-
- so since we know 3 y axis points on the graph and the axis of symmetry we can get another point without doing much work
- symmetry line = x = 1,
- calc'd x-intercept 0,-8
- the symmetry line is 1 and the known point is 0 since 1-0 = 1 we can add that to the x coordinate of y and keep the same y coordinate to get the mirrored point making another point on the graph (2,-8)
- since we need one more point for the graph we can choose say x=3, | x^2 -2x - 8
- y = 3^2 -3*2 - 8 = -5
- soooo the new point is (3,-5) if we mirror that along 1,-9 we get (-1, -5 ) because 3 is 2 more than 1, and 2 less than 1 is -1. we also keep the same y coordinate
- y = 3^2 -3*2 - 8 = -5
- so all points are:
- (1, -9)
- (0,-8)
- (2, -8)
- (3,-5)
- (-1,-5)
Example 6: Graph y = 2x^2 - 8x
• Find the vertex and the axis of symmetry. Sketch these in.
• Find the x-intercept by plugging in 0 for y.
• Find the y-intercept by plugging in 0 for x.
• Reflect your points across the axis of symmetry and connect your dots with a smooth
U-shaped (not V-shaped) curve.
Example 7: h = 16t^2 + 72t + 520
Suppose a particular “star” is projected from a firework at a starting height of 520 feet with an initial upward velocity of 72 ft/sec.
The equation:
h = 16t^2 + 72t + 520
gives the star’s height h in feet at time t in seconds.
a) How long will it take for the star to reach its maximum height?
b) What is the maximum height?