WRITE THE QUADRATIC EQUATION IN STANDARD FORM

Write the equation in standard form and find the axis of symmetry and direction of opening.

Problem 1 :

y = 2x² - 12x + 19

Problem 2 :

y = (1/2)x² + 3x + 1/2

Problem 3 :

y = -3x² - 12x - 7

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 ?

Answer Key

y = 2(x - 3)² + 1

The parabola is symmetric about y-axis and it is opening upward.

y = (1/2)(x + 3)² - 4

The parabola is symmetric about y-axis and it is opening upward.

y = -3(x + 2)² + 5

The parabola is symmetric about y-axis and it is opening down.

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-intercept is 1

b) the y-intercepts are -1 and -1/3.

Vertex is at (-1/3, -2/3)

Equation of axis of symmetry x = -2/3

d) Vertex is at (-1/3, -2/3).