uchihaitachi wrote:
Q. What is the minimum value of the function \(f(x) = 3x^2 + 6x +4\)?
A. -4
B. 1
C. 0
D. 4
E. 6
Solution:We see that the function is a quadratic function with a positive coefficient of x^2. Thus, its graph is an up-opening parabola with its vertex as its lowest point. Therefore, the y-value of the vertex is the minimum value of the function.
The x-value of the vertex can be obtained using the formula: x = -b/(2a), where a and b are the coefficients of x^2 and x, respectively. Therefore, the x-value of the vertex is -6/(2(3)) = -1, and the y-value of the vertex, i.e., the minimum value of the function, is f(-1) = 3(-1)^2 + 6(-1) + 4 = 3 - 6 + 4 = 1.
Alternate Solution:Alternatively, we can re-express the given quadratic function in vertex form, i.e. in the form f(x) = a(x - h)^2 + k. In this form, the coordinates of the vertex will be given by (h, k), and the minimum value of the function will be k. Setting 3x^2 + 6x + 4 equal to a(x - h)^2 + k, we obtain:
a(x - h)^2 + k = 3x^2 + 6x + 4
a(x^2 - 2xh + h^2) + k = 3x^2 + 6x + 4
ax^2 - 2ahx +ah^2 + k = 3x^2 + 6x + 4
Equating the coefficients of the quadratic terms, we get a = 3. Equating the coefficients of the linear terms and substituting a = 3, we get:
-2(3)hx = 6x
-6h = 6
h = -1
Finally, equating the constant terms, we obtain:
ah^2 + k = 4
3(-1)^2 + k = 4
3 + k = 4
k = 1
So, the vertex is (-1, 1), and the minimum value of the function is 1.
Answer: B
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