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What are the factors of 8a?
They are: 1, 2, 4, 8, a, 2a, 4a and 8a.
How do you write an equation for a parabola with a vertex at -4 2 and y-intercept -2?
1. The y-intercept of a parabola with equation y = ax^2 + bx + c is c. So, c = -22. The vertex is (x, y) = (-4, 2), where x = - b/2ax = - b/2a-4 = - b/2a(-4)(-2a) = (-b/2a)(-2a)8a = bSo we have:y = ax^2 + bx + c (substitute what you know: 2 for y, -4 for x, 8a for b, and -2 for c)2 = a(-4)^2 + (8a)(-4) + (-2)2 = 16a - 32a - 22 = - 16a - 2 (add 2 to both sides)4 = -16a (divide by -16 to both sides)-1/4 = aSince b = 8a, then b = 8(-1/4) = -2 Since a = -1/4, b = -2, and c = -2, then we can write the equation of the parabola asy = (-1/4)x^2 - 2x - 2.
What is -4 plus 10 plus 6a -2a plus 3?
-4 + 10 + 6a - 2a + 3 9 + 4a
Solve the equation 7x-8a equals 3x plus 24a for x?
7x - 8a = 3x + 24a First, add 8a to both sides: 7x = 3x + 24a + 8a Now subtract 3x from both sides to get all x terms on one side: 7x - 3x = 24a + 8a Simplify... 4x = 32a And divide both sides by 4: x = 32a / 4 = 8a x = 8a
What are the vertex and the axis of symmetry of the equation y equals 2x² plus 4x - 10?
In the form y = ax² + bx + c the axis of symmetry is given by the line x = -b/2a The axis of symmetry runs through the vertex, and the vertex is given by (-b/2a, -b²/4a + c). For y = 2x² + 4x - 10: → axis of symmetry is x = -4/(2×2) = -4/4 = -1 → vertex = (-1, -4²/(4×2) - 10) = (-1, -16/8 - 10) = (-1, -12)
