Respuesta :
You can use calculus to do this
try the first one
f'(x) = -x + 4 = 0 giving x = 4
f(4) = -8 + 16 - 11 = 8 -11 = -3 so it could be this one
- Oh no since the coefficient of x is negative it will have a maximum value
the correct one will be either C or D
Have you done any calculus?
try the first one
f'(x) = -x + 4 = 0 giving x = 4
f(4) = -8 + 16 - 11 = 8 -11 = -3 so it could be this one
- Oh no since the coefficient of x is negative it will have a maximum value
the correct one will be either C or D
Have you done any calculus?
- To solve this question, it is needed to find the vertex of quadratic equations.
Doing this, we get that the function is: f(x) =1/2x2 – 4x + 5, given by the third option.
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
If a > 0, the vertex is a minimum point, that is, the minimum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
Of the four options:
- The first two do not have minimums, as the a coefficient is negative, so they are not the answer.
Third equation:
[tex]f(x) = \frac{1}{2}x^2 - 4x + 5[/tex]
Has: [tex]a = \frac{1}{2}, b = -4, c = 5[/tex]
The x-value at the vertex is:
[tex]x_v = -\frac{b}{2a} = -\frac{-4}{2\frac{1}{2}} = -\frac{-4}{1} = 4[/tex]
The y-value at the vertex is:
[tex]\Delta = b^2-4ac = (-4)^2 - 4(\frac{1}{2})(5) = 16 - 10 = 6[/tex]
[tex]y_v = -\frac{\Delta}{4a} = -\frac{6}{4\frac{1}{2}} = -\frac{6}{2} = -3[/tex]
Thus, the minimum is located at (4,-3), and this is the correct answer.
A similar question is found at https://brainly.com/question/23559046
