WRITE THE QUADRATIC EQUATION IN STANDARD FORM
Every quadratic function in general form will be
ax2 + bx + c
The given quadratic function should be considered as
y = ax2 + bx + c
To convert the quadratic function into standard form, we have to rewrite the function as
y = a(x - h)2 + k
Using the method of completing the square, we can rewrite the quadratic function into standard form.
- Factor the leading coefficient of x2 and x.
- Write the coefficient of x, write it as a multiple of 2.
- Now the first two terms will look like a2 + 2ab or a2 - 2ab, inorder to complete the formula we have to add b2. Since we are adding this b2 inorder to complete the formula, to balance this, we have to subtract b2.
- a2 + 2ab + b2 = (a + b)2, a2 - 2ab + b2 = (a - b)2
By comparing the standard form of the quadratic function with the functions given below, we know that the parabola is symmetric about which axis and the direction of opening.
|
(y - k)2 = 4a(x - h) |
Open right |
|
(y - k)2 = -4a(x - h) |
Open left |
|
(x - h)2 = 4a(y - k) |
Open up |
|
(x - h)2 = -4a(y - k) |
Open down |
Write the equation in standard form and find the axis of symmetry and direction of opening.
Problem 1 :
y = 2x2 - 12x + 19
Solution :
y = 2x2 - 12x + 19
y = 2(x2 - 6x) + 19
y = 2[x2 - 2x(3) + 32 - 32] + 19
Here x2 - 2x(3) + 32 looks like a2 - 2ab + b2, we can rewrite it as (a - b)2
y = 2[x2 - 2x(3) + 32 - 32] + 19
= 2[(x - 3)2 - 32] + 19
= 2[(x - 3)2 - 9] + 19
Distributing 2, we get
= 2(x - 3)2 - 18 + 19
y = 2(x - 3)2 + 1
So, the standard form of the quadratic function is
y = 2(x - 3)2 + 1
The parabola is symmetric about y-axis and it is opening upward.
Problem 2 :
y = (1/2)x2 + 3x + 1/2
Solution :
y = (1/2)x2 + 3x + 1/2
y = (1/2)(x2 + 6x) + 1/2
y = (1/2)(x2 + 2x(3) + 32 - 32) + 1/2
y = (1/2)[x2 + 2x(3) + 32 - 32] + 1/2
Here x2 - 2x(3) + 32 looks like a2 - 2ab + b2, we can rewrite it as (a - b)2
y = (1/2)[(x + 3)2 - 9] + 1/2
y = (1/2)(x + 3)2 - (9/2) + 1/2
y = (1/2)(x + 3)2 - 8/2
y = (1/2)(x + 3)2 - 4
So, the standard form of the quadratic function is
y = (1/2)(x + 3)2 - 4
The parabola is symmetric about y-axis and it is opening upward.
Problem 3 :
y = -3x2 - 12x - 7
Solution :
y = -3x2 - 12x - 7
= -3[x2 + 4x] - 7
= -3[x2 + 2x(2) + 22 - 22] - 7
= -3[(x + 2)2 - 4] - 7
= -3(x + 2)2 + 12 - 7
y = -3(x + 2)2 + 5
So, the standard form of the quadratic function is
y = -3(x + 2)2 + 5
The parabola is symmetric about y-axis and it is opening down.
Problem 4 :
Use the equation x = 3y2 + 4y + 1
a) Find the x-intercepts
b) Find the y-intercepts
c) Write the equation of axis of symmetry
d) What are the coordinates of vertex ?
Solution :
x = 3y2 + 4y + 1
x = 3[y2 + (4/3)y] + 1
x = 3[y2 + 2 y (2/3) + (2/3)2 - (2/3)2] + 1
x = 3[(y + 2/3)2 - (4/9)] + 1
x = 3(y + 2/3)2 - (4/3) + 1
x = 3(y + 2/3)2 - (1/3)
a) x-intercepts :
To find x-intercepts, we have to apply y = 0
x = 3(0 + 2/3)2 - (1/3)
x = 3(4/9) - 1/3
x = (4/3) - (1/3)
x = 1
So, the x-intercept is 1
a) y-intercepts :
To find y-intercepts, we have to apply x = 0
0 = 3(y + 2/3)2 - (1/3)
3(y + 2/3)2 = 1/3
(y + 2/3)2 = 1/9
(y + 2/3) = 1/3 and -1/3
|
y + 2/3 = 1/3 y = 1/3 - 2/3 y = -1/3 |
y + 2/3 = -1/3 y = -1/3 - 2/3 y = -1 |
So, the y-intercepts are -1 and -1/3.
c) x = 3(y + 2/3)2 - (1/3)
From the standard form of the quadratic function, we know that the parabola is symmetric about x-axis and open rightward.
Vertex is at (-1/3, -2/3)
Equation of axis of symmetry x = -2/3
d) Vertex is at (-1/3, -2/3).
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