Find the vertex and y-intercept of the quadratic function!?

Without the equation of the function i can't verify your answers.

i'm going to assume that you're not in calculus, based on your description of how you're finding the required points.

Algebraically, you can find the vertex by completing the square:

y = ax^2 + bx + c

y = x^2 + (b/a)x + c/a

--> (x^2 + (b/a)x + (b^2/4a^2)) + c - b^2/4a^2

--> (x - (b/2a))^2 + c - b^2/4a^2

This is why the x-coordinate of the vertex is -b/2a. For example, if at this point it were (x-2)^2, the x-coordinate would be 2. (A simple way to think about this is that the vertex occurs when the part that is squared is equal to zero.)

Once you have the x-coordinate, simply plug this into the original equation for x, and solve it for y. This will be the y-coordinate of the vertex.

There are 2 information you would use: First bear in mind that the x-coordinate of the vertex is given by -b/(2a), for a quadratic in familiar kind (y=ax^2+bx+c). the following a=a million, b=2 and c=-3. nevertheless, the x-coordinate of the vertex is midway between the x-coordinates of the x-intercepts. both way you should get a similar answer. to locate the y coordinate of the vertex, change the x coordinate you cutting-edge in for x contained in the unique equation.