What is the minimum possible value for y=x^2+12x+5? I am not sure where to even start. Thank you!
the minimum point on a parabola is its vertex. So, find the vertex.
For y = ax^2+bx+c, the vertex is at x = -b/2a. In this case, that is
x = -12/2 = -6
AT x=-6, y = -31
This can be seen by rearranging the equation into the vertex form.
x^2+12x+5
= (x^2+12x+36) + 5 - 36
= (x+6)^2 - 31
Calculus is amazing. Take the derivative of the quadratic and get 2x+12. Set that equal to 0 to get x=-6. Plug that back in to the original equation to get y=-31
Ah, the mysterious realm of quadratic equations! Don't worry, I'm here to help you unravel this mathematical enigma with a touch of humor.
To find the minimum possible value for the given quadratic equation, we can employ a nifty trick. You see, quadratic equations take the shape of a parabola, which either opens upwards or downwards. In this case, since the coefficient of the x^2 term is positive (1 in this case), the parabola opens upwards, like a happy rainbow.
Now, the minimum value of a happy rainbow-looking parabola occurs at its vertex. The x-coordinate of the vertex can be found using the formula -b/2a, where a is the coefficient of the x^2 term (1) and b is the coefficient of the x term (12). Plugging in the values, we get x = -12/2(1) = -6.
To find the corresponding y-coordinate, we substitute this x-value back into the equation: y = (-6)^2 + 12(-6) + 5. Simplifying that, we get y = 36 - 72 + 5 = -31.
So, the minimum value for y in this case is -31. Voila! Maths and humor go hand in hand, don't they?
To find the minimum value of the quadratic function y = x^2 + 12x + 5, you can use a technique called completing the square. Here are the step-by-step instructions:
Step 1: Rewrite the given quadratic function in the form (x + h)^2 + k.
y = x^2 + 12x + 5
Start by adding and subtracting the square of half the coefficient of x (12/2 = 6) inside the parentheses:
y = x^2 + 12x + 36 - 36 + 5
Step 2: Group the perfect square trinomial (x + h)^2 and the constant term separately.
y = (x + 6)^2 + 5 - 36
Step 3: Simplify the equation.
y = (x + 6)^2 - 31
Step 4: Identify the vertex of the parabola.
The vertex of the parabola is given by the values (h, k) from the equation (x + h)^2 + k.
In this case, the vertex is (-6, -31).
Since the vertex of the parabola lies at the minimum point, the minimum value of y is equal to the y-coordinate of the vertex. Therefore, the minimum possible value for y = x^2 + 12x + 5 is -31.
To find the minimum possible value of the quadratic equation y = x^2 + 12x + 5, you can use a concept called completing the square. This involves rewriting the equation in a specific form, which will allow you to determine the vertex of the parabola and, therefore, the minimum value of the equation.
Here are the steps:
1. Start with the equation y = x^2 + 12x + 5.
2. To complete the square, add and subtract the square of half the coefficient of x (which is (12/2)^2 = 36) within the equation:
y = x^2 + 12x + 36 - 36 + 5.
Now, the equation can be rewritten as:
y = (x^2 + 12x + 36) - 31.
3. Simplify the equation inside the parentheses, which is a perfect square trinomial: (x + 6)^2.
The equation now becomes:
y = (x + 6)^2 - 31.
4. Since (x + 6)^2 is always non-negative, the minimum value will occur when it is equal to zero. Thus, y is minimized when (x + 6)^2 = 0.
Therefore, the minimum possible value of y is -31, which is the constant term (-31) obtained in step 2.