The minimum of a parabola is located at (-1,-3). the point (0,1) is also on the graph. which equation can be solved to determine the a value in the function representing the parabola?

The equation that can be used to solve for a is 1 = a(0=1)²-3.

Explanation:

In this case, I would model the parabola in vertex form.  Vertex form is y = a(x-h)²+k, where (h,k) is the vertex of the parabola.  Using the information from the question, we can plug in values to get 1 = a(0+1)²-3.  (This could be simplified as 1 = a(1)²-3 ⇒ 1 = a-3 ⇒ a = 4, but we are only interested in finding the equation that can solve for a.)

The correct answer is A, 1 = a(0 + 1)^2 – 3

Step-by-step explanation:

The verified guy made a typo, putting = instead of + :P

no solution

step-by-step explanation:

there is not enough information in the problem to solve.

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.