Quadratic Functions Flashcards
(22 cards)
what is a parabola
the graph of the quadratic function
y = ax² + bx + c
what determines whether the parabola opens upwards or downwards
a, the coefficient of x²
what is the turning point
the minimum or maximum point of a quadratic curve
turning point of parabola of a>0
it is a minimum point
turning point of parabola of a<0
it is a maximum point
general form of quadratic functions
y = ax² + bx + c
factorised form of quadratic functions
y = a(x - p)(x - q)
(i) factorise 2x² - 12x + 10 and state the coordinates of the turning point of the graph of y = 2x² - 12x + 10
(ii) hence, sketch the graph and state the minimum value of the function y = 2x² - 12x + 10
(i)
y = 2x² - 12x + 10
= 2(x² - 6x + 5)
= 2(x - 1)(x - 5)
x= 1 or x= 5
At the turning point,
x = (1+5)/2
= 3
y= 2(3 - 1)(3 - 5)
= -8
∴ the coordinates of the turning point are (3, -8)
(ii)
Sub x = 0 into y = 2x² - 12x + 10
When x = 0, y = 10
∴ the graph cuts the y- axis at (0,10)
Sub y = 0,
2(x - 1)(x - 5) = 0
x - 1 = 0 or x - 5 = 0
x = 1 or 5
∴ the graph cuts the x- axis at (1, 0) and (5, 0)
completing the square form
a(x - h)² + k
what points do you need to sketch curve
points of intersection of the curve with the axes and the turning point
complete the square of 2x² - 6x + 5
2x² - 6x + 5 = 2(x² - 3x + 5/2)
= 2[x² - 3x + (-3/2)² - (-3/2)² + 5/2]
= 2[(x - 3/2)² - 9/4 + 10/4]
= 2[(x - 3/2)² + 1/4]
= 2(x - 3/2)² + 1/2
Explain why 2(x - 3)² - 8 can never be less than -8
Since (x - 3)² ≥ 0 for all values of x, then 2(x - 3)² - 8 ≥ -8
∴ 2(x - 3)² - 8 can never be less than -8
complete the square of -4x² + 2.8x - 0.5
-4(x - 0.35)² + 0.99
turning point/vertex coordinates of graph of y = a(x - h)² + k
(h, k)
find the coordinates of the turning point using y = 3(x-2)² - 27
method 1:
Let (x-2)² = 0
Then x = 2 and f(2) = -27
turning point (2, -27
method 2:
3(x-2)² - 27 is in the form a(x - h)² + k
The coordinates are (h, k) = (2, -27)
how to find y intercept
sub x = 0 into y = ax² + bx + c
how to find x intercept
sub y = 0 into y = ax² + bx + c
2 conditions for the curve y = ax² + bx + c to lie completely above the x-axis
- a>0
- minimum value of ax² + bx + c is positive
2 conditions for the curve y = ax² + bx + c to lie completely below the x-axis
- a<0
- maximum value of ax² + bx + c is negative
Does y= -3x² - 12x - 27 lie completely above or below the x-axis? Explain your answer.
y = -3x² - 12x - 27
= -3(x² + 4x + 9)
= -3(x² + 4x + 2² - 2² + 9)
= -3[(x² + 2)² + 5]
= -3(x + 2) - 15
Since a = -3 which is < 0 and the
maximum value of -3x² - 12x - 27 is negative, then the curve lies completely below the x-axis.
Does y= 1/2 x² - 4x + 9 lie completely above or below the x-axis? Explain your answer.
y = 1/2 x² - 4x + 9
= 1/2(x² - 8x + 18)
= 1/2[x² - 8x + (-4)² - (-4)² + 18]
= 1/2[(x - 4)² + 2]
= 1/2(x - 4)² + 1
Since a = 1/2 which is > 0 and the
minimum value of 1/2 x² - 4x + 9 is positive, then the curve lies completely above the x-axis.
The height, y metres, of a ball hit by a bat can be modelled by the equation
y = -0.04x² + 2x + 24, where x is the horizontal distance travelled by the ball in metres.
Find
(i) the greatest height reached by the ball and the corresponding horizontal distance travelled,
(ii) the horizontal distance travelled by the ball before it hits the ground.
(i)
-0.04x² + 2x + 24
= -0.04(x² - 50x - 600)
= -0.04[x² - 50x + (-25)² - (-25)² - 600]
= -0.04[(x - 25)² - 625 - 600]
= -0.04[(x - 25)² - 1225]
= -0.04(x - 25)² + 49
∴ the greatest height reached by the ball is 49 m and the corresponding horizontal distance travelled is 25 m.
(ii) When y = 0,
-0.04x² + 2x + 24 = 0
-0.04(x - 25)² + 49 = 0
0.04(x - 25)² = 49
(x - 25)² = 1225
x - 25 = ± √1225
= ± 35
x = 25 ± 35
= -10 (rej as distance > 0) or 60
∴ the ball travels a horizontal distance of 60 m before it hits the ground.
