I have to find the standard form of a quadratic function with the vertex of -1,4 and x intercepts of -3,0 and 1,0. What do I do when I have two x intercepts like this?
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When you have two x-intercepts of a quadratic function, you can use the factored form of the quadratic equation to find the standard form. The standard form of a quadratic function is generally written as ax^2 + bx + c = 0, where a, b, and c are constants.
To find the standard form, you need to start by finding the equation in factored form using the given x-intercepts. In this case, the x-intercepts are -3 and 1, which means that the quadratic function can be factored as (x + 3)(x - 1) = 0.
Next, you can expand the equation by multiplying out the two expressions in the factored form. In this case, you would have (x + 3)(x - 1) = x^2 + 2x - 3 = 0.
Now, you have the quadratic equation in standard form, where a = 1, b = 2, and c = -3. Therefore, the standard form of the quadratic function with the given properties is x^2 + 2x - 3 = 0.
Remember that the vertex of the quadratic function is (-1, 4). To verify that the vertex is correct, you can use the formula for the vertex of a quadratic function, which is given by (-b/2a, f(-b/2a)), where b and a are the coefficients of x in the standard form. In this case, substituting the values a = 1 and b = 2 into the formula, you would get (-2/2(1), f(-2/2(1))). Simplifying further, you would have (-1, f(-1)), which indeed matches with the given vertex (-1, 4).