Let f(x) = 3x^2 + bx + c. find values b and c such that the vertex is (4/3, -10/3)

From calculus we know that the vertex of a quadratic function is its one and only critical point.

f(x) = 3x^2 + bx + c

df/dx = 6x+b is not undefined anywhere so,

6x+b = 0

x = -b/6 (Which you might recall from algebra as x = -b/2a) is the x coordinate. So we assume that equal the x coordinate of our vertex : -b/6 = 4/3  ===(Multiplying both sides by -6)> b = -8

So our quadratic function takes the form,

f(x) = 3x^2 - 8x + c

To find c just plug in 4/3 for x and assume it is -10/3:

3(4/3) ^ 2 - 8*(4/3) + c = -10/3

Solving for c we get: c = 2 [If I did my arithmetic correctly that is]

f(x) = 3x^2 - 8x +2

Comment if you have any questions or you think I made a mistake somewhere.

Hope this helps :)

false

step-by-step explanation:

false.   two legs of the trapezoid would be congruent, but the other two would not.   were all the legs congruent, then we'd have a square; were the 2 shorter sides congruent and the 2 longer sides also congruent, we'd have a rectangle.