Question

please help me find the vertex and intercepts for y=2x^2-3+3

Answer 1

Is that really -3+3? Or isn't that 3x?

Answer 2

sorry, it is y =2x^2-3x+3

Answer 3

If that's the case, we can get the y-int by substituting 0 to x and the x-int by substituting 0 to y this time.

y-int: y = 2(0) - 3(0) +3; y=3
x-int: 0 = 2x^2 - 3x + 3

The x-intercept is not factorable, but if you use the quadratic formula, you will see that the x-intercepts are imaginary and not in the real number line. Therefore, there aren't any x-ints.

For the vertex, the x-coordinate is known to be (-b)/(2a). So x-coor: -(-3)/2(2). The x-coor is 3/4. For the y-coordinate, we substitute 3/4 to the function, since the vertex IS lying on the curve and is, thus, in the domain of the function. y= 2(3/4)^2 - 3(3/4) + 3. The vertex would then be at (3/4, f(3/4)). f(3/4) is also the same as substituting 3/4 to the function. :)