Consider the quadratic function
f(x) = x^2 + 12x + 40. Find the vertex.
complete the square,
f(x) = x^2 + 12x + 40
= x^2 + 12x + 36 - 36 + 40
= (x+6)^2 + 4
so the vertex is (-6,4)
or
the x co-ordinate of the vertex for
f(x) = ax^2 + bx + c is -b/(2a)
so x = -12/2 = -6
sub x=-6 back into the function to find f(-6) = 36 - 72 + 40 = 4
or, if you know Calculus,
f'(x) = 2x + 12 = 0 at the vertex
2x + 12 = 0
x = -6 etc
Thank you! I got many more of these to do, so just wanted to make sure my first one was correct! Thanks!
To find the vertex of a quadratic function in the form f(x) = ax^2 + bx + c, you can use the formula:
x = -b / (2a)
In this case, the function is f(x) = x^2 + 12x + 40. So, a = 1, b = 12, and c = 40.
Using the formula, we can find the x-coordinate of the vertex:
x = -b / (2a)
= -12 / (2 * 1)
= -12 / 2
= -6
Now we can substitute this value of x into the original function to find the y-coordinate:
f(x) = (x^2) + 12x + 40
= (-6)^2 + 12(-6) + 40
= 36 - 72 + 40
= 4
Therefore, the vertex of the quadratic function f(x) = x^2 + 12x + 40 is (-6, 4).
To find the vertex of a quadratic function of the form f(x) = ax^2 + bx + c, we can use the formula:
x = -b / (2a)
In our case, a = 1, and b = 12. Plugging those values into the formula, we get:
x = -12 / (2 * 1)
x = -12 / 2
x = -6
Now that we have the x-coordinate of the vertex, we can substitute it back into the original equation to find the y-coordinate. Plugging x = -6 into the equation f(x) = x^2 + 12x + 40, we get:
f(-6) = (-6)^2 + 12(-6) + 40
f(-6) = 36 - 72 + 40
f(-6) = 4
Therefore, the vertex of the quadratic function f(x) = x^2 + 12x + 40 is (-6, 4).