Write a quadratic function in vertex form 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 ANSWER A quadratic function for the parabola is y = (x – 1)2 – 2.

Write a quadratic function in intercept form 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.

Write a quadratic function in intercept form 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. 1 2 – = a Solve for a. A quadratic function for the parabola is 1 2 – (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.

Write a quadratic function in standard form 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 STEP 2 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.

Write a quadratic function in standard form 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 3 Solve 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 ANSWER 2. vertex: (–3, 1) passes through: (0, –8) ANSWER y = (x + 3)2 + 1 y = (x + 2)(x – 5) 1 4 ANSWER 3. x-intercepts: –2, 5 passes through: (6, 2)

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 ANSWER 5. (–2, –1), (0, 3), (4, 1) –5 12 7 6 ANSWER y = x2 + x + 3. 6. (–1, 0), (1, –2), (2, –15) y = 4x2  x + 3 ANSWER