How to Graph a Quadratic Function by HandUsing Its PropertiesGraph f1x2 = - 3x2 + 6x + 1 using its properties. Determine the domain and therange of f. Determine where f is increasing and where it is decreasing. EXAMPLE 3 Step-by-Step SolutionStep 1: Determine whether thegraph of f opens up or down.Step 2: Determine the vertex andaxis of symmetry of the graph of f.Step 3: Determine the intercepts ofthe graph of f. In Example 2, it was determined that the graph of f1x2 = - 3x2 + 6x + 1 opensdown because a = - 3 6 0.In Example 2, the vertex was found to be at the point whose coordinates are (1, 4).The axis of symmetry is the line x = 1.The y-intercept is found by letting x = 0. The y-intercept is f102 = 1. Thex-intercepts are found by solving the equation f1x2 = 0. f1x2 = 0- 3x2 + 6x + 1 = 0 a = −3, b = 6, c = 1 The discriminant b2 - 4ac = 162 2 - 41 - 32 112 = 36 + 12 = 48 7 0, so theequation has two real solutions and the graph has two x-intercepts. Use thequadratic formula to find thatx = - b + 2b2 - 4ac 2a = - 6 + 248 21 - 32 = - 6 + 423- 6 ≈ - 0.15 andx = - b - 2b2 - 4ac 2a = - 6 - 248 21 - 32 = - 6 - 423- 6 ≈ 2.15 The x-intercepts are approximately - 0.15 and 2.15.
How to Graph a Quadratic Function by Hand
Using Its Properties
Graph f1x2 = - 3x2 + 6x + 1 using its properties. Determine the domain and the
range of f. Determine where f is increasing and where it is decreasing.
EXAMPLE 3
Step-by-Step Solution
Step 1: Determine whether the
graph of f opens up or down.
Step 2: Determine the vertex and
axis of symmetry of the graph of f.
Step 3: Determine the intercepts of
the graph of f.
In Example 2, it was determined that the graph of f1x2 = - 3x2 + 6x + 1 opens
down because a = - 3 6 0.
In Example 2, the vertex was found to be at the point whose coordinates are (1, 4).
The axis of symmetry is the line x = 1.
The y-intercept is found by letting x = 0. The y-intercept is f102 = 1. The
x-intercepts are found by solving the equation f1x2 = 0.
f1x2 = 0
- 3x2 + 6x + 1 = 0 a = −3, b = 6, c = 1
The discriminant b2 - 4ac = 162 2 - 41 - 32 112 = 36 + 12 = 48 7 0, so the
equation has two real solutions and the graph has two x-intercepts. Use the
quadratic formula to find that
x = - b + 2b2 - 4ac
2a = - 6 + 248
21 - 32 = - 6 + 423
- 6 ≈ - 0.15
and
x = - b - 2b2 - 4ac
2a = - 6 - 248
21 - 32 = - 6 - 423
- 6 ≈ 2.15
The x-intercepts are approximately - 0.15 and 2.15.
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