Did you mean f(x)=x2-2x+5 ? If so the x value of the vertex is found by using the formula x=-b/2a where a is the coefficient of x^2 and b is the coeffiecient of x x=-(-3)/(2*-2) =3/4 y=-2*(3/4)^2-3*(3/4)+7
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Since this question is in the calculus section, I'm assuming you know how to take the derivative. We know that y = -2x2 + 2x + 3 is a parabola, so it has one vertex, which is a minimum. We can use the first derivative test to find this extreme point.First, take the derivative:y' = -4x + 2Next, set y' equal to zero:0 = -4x + 2Then solve for x:4x = 2x = 2This is the x-coordinate of the vertex. To find the y-coordinate, plug x = 2 back into the original equation:y = -2x2 + 2x + 3y = -8 + 4 + 3y = -1So the vertex is at (2, -1).
Complete the square to find the transformation. -2(x-2)^2-4 the y value is -4. so the vertex is at (2,-4) OR Use the vertex formula: -b/2a a=-2 b=8 x=-8/(2)-2 x=2 Run 2 through the original equation: y=-(2)^2+8(2)-12 y=4
2x2-2 = -5x-1 2x2+5x-2+1 = 0 2x2+5x-1 = 0 Use the quadratic equation formula to find the values of x: x = 0.1861406616 or x = -2.686140662
Use a calculator.
Given three vertices, the two that are the furthest apart lie at the ends of a diagonal. Reflect the square in this diagonal. The third vertex will be where the missing vertex should be.