Function f is a quadratic function. The graph of y = f(x) in the xy-plane has a vertex at (6, 10) contains the point (5, 8), and has a y-intercept at (0, a) The graph of y = 4f(x) has a y-intercept at (0, b) What is the positive difference between a and b?

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Function f is a quadratic function. The graph of y = f(x) in the xy-plane has a vertex at (6, 10) contains the point (5, 8), and has a y-intercept at (0, a) The graph of y = 4f(x) has a y-intercept at (0, b) What is the positive difference between a and b?

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Vertex Form Equation: Vertex form of quadratic function: $ y = a(x - h)^2 + k $ Given vertex: (6, 10) Equation: $ y = a(x - 6)^2 + 10 $Substitute Point: Point (5, 8) lies on the parabola. Substitute (5, 8) into the equation: $ 8 = a(5 - 6)^2 + 10 $ $ 8 = a(1)^2 + 10 $ $ 8 = a + 10 $ Solve for a: $ a = -2 $Find Y-Intercept: Equation of the parabola: $ y = -2(x - 6)^2 + 10 $ Find y-intercept (0, a):  Substitute x = 0: $ y = -2(0 - 6)^2 + 10 $ $ y = -2(36) + 10 $ $ y = -72 + 10 $ $ y = -62 $ So, a = -62Graph Transformation: Graph of $ y = 4f(x) $ New equation: $ y = 4[-2(x - 6)^2 + 10] $ $ y = -8(x - 6)^2 + 40 $ Find y-intercept (0, b): Substitute x = 0: $ y = -8(0 - 6)^2 + 40 $ $ y = -8(36) + 40 $ $ y = -288 + 40 $ $ y = -248 $ So, b = -248Calculate Positive Difference: Positive difference between a and b: $ |a - b| = |-62 - (-248)| $ $ |a - b| = |-62 + 248| $ $ |a - b| = |186| $ Positive difference = 186

Function $f$ is a quadratic function. The graph of $y = f(x)$ in the $xy$ -plane has a vertex at $(6, 10)$ , contains the point $(5, 8)$ , and has a $y$ -intercept at $(0, a)$ . The graph of $y = 4f(x)$ has a $y$ -intercept at $(0, b)$ . What is the positive difference between $a$ and $b$ ?

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