Find the vertex
y = x2 + 2
(0, 2)
y = 2x2 - 5x - 11
Standard form
Vertex (1, 2)
Passes through (2, 3)
y = (x - 1)2 + 2
A ball is thrown along the path y = -1/4(x - 3)2 + 4 What is the maximum height the ball reaches?
Maximum at the vertex, (3, 4). Max height = 4.
y = 1/2x2 Plug in -3 for x:
y = 1/2(-3)2
y = 1/2(-9)
y = -4.5
(-3)2 should be 9.
y = (x - 3)2 - 6
(3, -6)
y = (x - 4)(x + 1/2)
The graph of y = x2, except opening downard
y = -x2
Does the graph of this quadratic have a maximum point or a minimum point?
y = 0.06x2 - 1.4x + 3.7
Opens upward (a = 0.06, which is > 0) so it has a minimum point at the vertex.
y =-3(x - 1)(x + 3) Plug in -2 for x.
y = -3(-2 - 1)(-2 + 3)
y = -3(-3)(1)
y = -6.
(-3)(-3)(1) = 9
y = (x - 1)(x + 3)
(-1, -4)
y = -1/3(x + 0.5)2 - 3.7
Vertex form
x intercepts at (3, 0) and (5, 0) and passing through point (4, 1)
y = -(x - 3)(x - 5)
2Some quadratic data is processed through computer software (like Desmos) and you get the equation
y = 2x2 - 6x -1, with an R2 value of 0.3961. Would you consider this strongly correlated or weakly correlated? Explain your answer.
Weak, because the R2 value is fairly low. Strong correlations give R2 numbers like .89 or .95.
Plug in the x intercepts (-2, 0) and (8, 0) to the intercept form equation:
y = a(x - 2)(x + 8)
Should read y = a(x + 2)(x - 8)
y = 2x2 - 8x + 3
(2, -5)
y = (x - 5)2
Vertex form (or intercept form)
Vertex at (-3, -5) and passing through point (0, 2)
y = 7/9(x + 3)2 - 5
A ball is thrown into the air and follows a path given by the equation
y = -1/4x2 + 6x + 1.
What is the maximum height the ball reaches?
Maximum at the vertex (12, 37). 37 feet.
What is the vertex point of y = -2/3(x + 7)2 - 5?
(7, -5)
Vertex should be (-7, 5)
y = -1/2x2 - 4x - 1/2
(-4, 7.5)
x = a(y - k)2 + h
Vertex form of a sideways parabola
Same vertex as y = (x + 3)2 - 2 but passing through (1, -6)
y = -1/4(x + 3)2 - 2
A pitched baseball follows the parabolic path given by
y = -.0032(x - 25)2 + 7
Can this pitch be a strike? That is, when it travels 60 feet to home plate, is its height between 1.5 and 3.5 feet?
Yes, plugging in 60 for x gives 3.1 feet. The ball is in the strike zone.
Plug in (-4, 2) for (x, y) and solve for a.
y = a(x + 2)2 + 4
2 = a(-4 + 2) + 4 plug in
2 = a(-2) + 4 subtract 4
-2 = -2a divide by -2 and a = 1
Forgot to copy down the "2" for the square.