EXAMPLE 1 Write a quadratic function in vertex form
Write a quadratic function for the parabola shown.
SOLUTION
Use vertex form because the vertex is given.
y = a(x – h)2 + k Vertex form
y = a(x – 1)2 – 2 Substitute 1 for h and –2 for k.
Use the other given point, (3, 2), to find a.2 = a(3 – 1)2 – 2 Substitute 3 for x and 2 for y.
2 = 4a – 2 Simplify coefficient of a.
1 = a Solve for a.
EXAMPLE 1 Write a quadratic function in vertex form
A quadratic function for the parabola is y = (x – 1)2 – 2.
ANSWER
EXAMPLE 2 Write a quadratic function in intercept form
Write a quadratic function for the parabola shown.
SOLUTION
Use intercept form because the x-intercepts are given.
y = a(x – p)(x – q) Intercept form
y = a(x + 1)(x – 4) Substitute –1 for p and 4 for q.
EXAMPLE 2 Write a quadratic function in intercept form
Use the other given point, (3, 2), to find a.
2 = a(3 + 1)(3 – 4) Substitute 3 for x and 2 for y.
2 = –4a Simplify coefficient of a.
Solve for a.12
– = a
A quadratic function for the parabola is
12
– (x + 1)(x – 4) .y =
ANSWER
EXAMPLE 3 Write a quadratic function in standard form
Write a quadratic function in standard form for the parabola that passes through the points (–1, –3), (0, –4), and (2, 6).
SOLUTION
STEP 1
Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.
EXAMPLE 3 Write a quadratic function in standard form
–3 = a(–1)2 + b(–1) + c Substitute –1 for x and 23 for y.
–3 = a – b + c Equation 1
–3 = a(0)2 + b(0) + c Substitute 0 for x and –4 for y.
–4 = c Equation 2
6 = a(2)2 + b(2) + c Substitute 2 for x and 6 for y.
6 = 4a + 2b + c Equation 3
Rewrite the system of three equations in Step 1 as a system of two equations by substituting –4 for c in Equations 1 and 3.
STEP 2
EXAMPLE 3 Write a quadratic function in standard form
a – b + c = –3 Equation 1
a – b – 4 = –3 Substitute –4 for c.
a – b = 1 Revised Equation 1
4a + 2b + c = 6 Equation 3
4a + 2b – 4 = 6 Substitute –4 for c.
4a + 2b = 10 Revised Equation 3
STEP 3Solve the system consisting of revised Equations 1 and 3. Use the elimination method.
EXAMPLE 3 Write a quadratic function in standard form
a – b = 1 2a – 2b = 2
4a + 2b = 10 4a + 2b = 10
6a = 12
a = 2
So 2 – b = 1, which means b = 1.
The solution is a = 2, b = 1, and c = –4.
A quadratic function for the parabola is y = 2x2 + x – 4.
ANSWER
GUIDED PRACTICE for Examples 1, 2 and 3
Write a quadratic function whose graph has the given characteristics.
1. vertex: (4, –5)
passes through: (2, –1)y = (x – 4)2 – 5
ANSWER2. vertex: (–3, 1)
passes through: (0, –8)y = (x + 3)2 + 1
ANSWER
3. x-intercepts: –2, 5 passes through: (6, 2)
y = (x + 2)(x – 5)14ANSWER
GUIDED PRACTICE for Examples 1, 2 and 3
Write a quadratic function in standard form for the parabola that passes through the given points.
4. (–1, 5), (0, –1), (2, 11)
y = 4x2 – 2x – 1
y = x2 + x + 3. –512
76
5. (–2, –1), (0, 3), (4, 1)
6. (–1, 0), (1, –2), (2, –15)
y = 4x2 x + 3
ANSWER
ANSWER
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