For a quadratic function , the turning point would be the vertex of the parabola.
since your equation is written in vertex form
f(x) = (x+3)^2 - 16
the vertex is (-3,-16) , which would be your turning point.

If your quadratic is written in the form
f(x) = ax^2 + bx + c
there is a quick way to find the vertex
the x of the vertex is -b/(2a)
once you have that x, sub it into the equation to get the y
e.g.
y = 3x^2 + 9x - 11
the x of the vertex is -9/6 = -3/2
then y = 3(9/4) + 9(-3/2) - 11
= 24/4 - 54/4 - 44/4
= -37/2
the vertex or turning point is (-3/2 , -37/2)

a) Solve for x: 2x² + 5x = 3 b) Solve for x and y: x + y = 4 and x² + y² + xy - 12 =0 c) Solve for x: 8 2x + 3 = ½x² d) Factorise: 6x² - 7xy - 3y² e) Factorise: 4x³ + 5x²y - 4xy² - 5y³

help my daughter has a math problem and came to me for help, I am lost.... what is the gcf for the following trinomial? x^3-18x^2+28x I am assumming you are trying to factorise we can simply take out an x to give: x(x^2-18x+28)

Hi Just wandering When I factorise the following : 4s-16t+20r It will equal to 2 ( 2s - 8t + 10 r ) This was the answer in the book However , I managed to factorise it to become : 4 ( s-4t+5r) is this correct ? Because when you