What is the minimum possible value for y=x^2+12x+5? I am not sure where to start.
Assuming you are not allowed to use calculus, complete the square to find the vertex.
x^2 + 12 x = y-5
x^2 + 12 x + 6^2 = y - 5 + 6^2
(x+6)^2 = y + 31
vertex at x = -6, y = -31
so never smaller than -31
differentiate y with respect to x (dy/dx). That gives you the slope at any point on the graph. The slope is always 0 at the max and min points so put dy/dx = 0 and solve for x.
Once you get that value of x put it back into y to find the corresponding y value for the minimum point.
To find the minimum possible value of the quadratic function y = x^2 + 12x + 5, you can use the concept of vertex.
Step 1: The quadratic function is in the form of y = ax^2 + bx + c, where a = 1, b = 12, and c = 5.
Step 2: To find the x-coordinate of the vertex, you can use the formula x = -b/2a. Plugging in the values, x = -12/(2*1) = -6.
Step 3: To find the y-coordinate of the vertex, substitute the value of x back into the equation. y = (-6)^2 + 12(-6) + 5 = 36 - 72 + 5 = -31.
Step 4: The y-coordinate of the vertex is the minimum value of the quadratic function. Therefore, the minimum possible value for y = x^2 + 12x + 5 is -31.
So, the solution is -31.