find the vertex of a parabola of f(x)=3x^2+18x+10 quadratic function

Question

help with college algebra homework please

Answer 1

The x-coordinate of the vertex will be = -b/(2a) when you're given an equation of the form y = ax^2 + bx + c.

So here a = 3, b = 18, and so the x-coord is -18/(2*3) = -3.

Now just plug that into the equation to find the y-coord of the vertex:

3(-3)^2 + 18(-3) + 10 = -17

So the vertex is (-3,-17)

Answer 2

Another way to find the vertex would be to calculate the velocity at 0.  To find the velocity at 0 you take the derivative of the equation which is y=6x+18. set Y equal to zero and solve.

  0=6x+18.

-18=6x

x= -3. Then plug -3 back into the original equation of

f(x)=3x^2+18x+10

F(-3)= 3 (-3^2) + 18(-3)+10

F(-3)= -17

So the vertex is (-3,-17)