Which equation in standard form has a graph the passes through (1,-2) and has a of 2?

[tex]Which equation in standard form has a graph the passes through (1,-2) and has a of 2?[/tex]

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Which equation in standard form has a graph the passes through (1,-2) and has a of 2?

[tex]Which equation in standard form has a graph the passes through (1,-2) and has a of 2?[/tex]

1) The equation is given in factored form

2) The key characteristics of the parabola are;

a. The parabola has 2 real roots and extends to infinity on both sides

b. The x-intercepts of the parabola are (1/2, 0) and (-6, 0)

c. The y intercept is (0, -6)

d. The vertex point is (-2.75, -21.125)

Step-by-step explanation:

The given equation is f(x) = (2·x -1)(x + 6)

Therefore the equation is given in factored form

The key characteristic of the parabola revealed from the form f(x) = (2·x -1)(x + 6) are;

1) Writing the parabola in the intercept form, a(x - p)(x -q) we have;

f(x) = (2·x -1)(x + 6) = 2·(x - 1/2)(x + 6)

p = 1/2, q = -6

Therefore;

Given that a is positive the parabola is concave upwards

2) a. The parabola has 2 real roots and extends to infinity on both sides

b. The x-intercepts of the parabola are p and q which are (1/2, 0) and (-6, 0)

c. The y intercept is a·(-p)·(-q) = 2×(-1/2)×6 = -6

The y intercept is (0, -6)

d. Expanding we get;

f(x) = 2·x² + 11·x - 6

From the

The vertex is h = -b/(2·a) = -11/(2×2) = -2.75

h(2.75) = 2×(-2.75)² + 11×(-2.75) - 6 = -21.125

Therefore, the function is symmetrical about the vertex point (-2.75, -21.125)