Determine whether this quadratic function has a maximum value or a minimum value, and then find the value:?

Question

f(x) = 2x^2 + 8x + 5

maximum value; 4
Minimum value; -1
maximum value; 3
minimum value; -3

Answer 1

Look at the coefficient of the x^2 term. If it is positive, the parabola opens up and has a minimum value; if it is negative, the parabola opens down and has a maximum. You have 2x^2, so you will have a minimum value.

Next find the vertex. The x-coordinate of the vertex lies at -b/2a. So -8/(2)(2) = -8/4 = -2. Plug this value in to get the y-coordinate of the vertex. 2(-2^2) + (8)(-2) + 5 = 8 - 16 + 5 = -3.

So you have a minimum value of -3.

Answer 2

for y = ax^2 + bx +c then

if a is positive, the function has a minimum. If a is negative, the function has a maximum.

Differentiate the function. You get 2ax+b. The minimum or maximum is where the dervative is 0 because this is where the slope is zero. This is because if the slope is not zero, then moving to the left or the right along the curve would result in a higher or lower value, so the points where the slope is not zero can't be maximums or minimums.

Solve for x when the derivative 2ax+b is 0. You get x=-b/2a. This is the x position where the minimum or maximum occurs.

Plug x=-b/2a into the original equation and you get the value of y at the min or max, which is the minimum or maximum of the function.

Answer 3

It is a function, so it doesn't open to the left or to the right. The only possible choices are opening upward or downward:
PARABOLAE
(x - h)² = 4p(y - k) ---> opening upward
(x - h)² = -4p(y - k) ---> opening downward

When you convert this where y is a function then
y = ax² + bx + c, a > 0 ---> opening upward
y = ax² + bx + c, a < 0 ---> opening downward

Since in your example
f(x) = 2x² + 8x + 5
a > 0, so it opens upward. Since it opens upward, the vertex is at the bottom, and it is a minimum value. The vertex:

(x - h)² = 4p(y - k)
Vertex: V(h,k)

Trying to convert y = ax² + bx + c
y/a = x² + b/a x + c/a
y/a = x² + b/a x + b²/4a² - b²/4a² + c/a
y/a - c/a + b²/4a² = (x + b/2a)²
(x + b/2a)² = 1/a (y - c + b²/4a)
[x - (-b/2a)]² = 1/a [y - (c - b²/4a)]

Therefore, the vertex is at
(-b/2a,c - b²/4a)

Using(from given)
a = 2
b = 8
c = 5

thus,
c - b²/4a
= 5 - 8²/4(2)
= 5 - 16/8
= 5 - 8
= -3

Therefore, f(x) has a minimum value which is -3.

^_^

Answer 4

We know that the parabola opens up, because the coefficient for the x^2 is a positive number (2). That means that the function will have a minimum value but no maximum value (it will just continue up to infinity). I don't remember how we were taught in algebra to calculate minimum value, but in calculus, we took the derivative and set it equal to zero.
f'(x) = 4x + 8
0 = 4x + 8
The minimum value is at x = -2.
y = 2(-2)^2 + 8(-2) + 5 = -3

Answer 5

I'm not exactly sure what you are asking since you seem to have the answers here. But, if these are the answers and you want to know how to get them, then the best thing to do would be to first take the derivative. The place where the derivative is equal to zero will be your critical point (max or min).

f'(x)=4x+8

0=4x+8
4x=-8
x=-2

So, none of your answer choices are correct.

Taking the second derivative will tell you whether it is a max or min.

f''(x)=4

Since the second derivative is greater than zero, then you know that it is a min. You would also know that since the x^2 term is positive.

Answer 6

f(x) = 2x^2 + 8x + 5

x = (-b/2a)

x = (-8)/(2(2))
x = -8/4
x = -2

f(-2) = 2(-2)^2 + 8(-2) + 5
f(-2) = 2(4) - 16 + 5
f(-2) = 8 - 16 + 5
f(-2) = -8 + 5
f(-2) = -3

Minimum Value of -3

if the highest exponent has a coefficient with a negative, then you have a maximum value, if the highest exponent has a coefficient with a positive, then you have a minimum value.

Answer 7

for max and minimum values u have to find first derivative of
f(x) so,
f'(x)=4x+8+0
f'(x)=4x+8

now find the critical values i.e.
those values of x where f'(x) is zero

4x+8=0
x=-2

now decide whether for x=-2 is minima or maxima
find second derivative of f(x)
so, f''(x)=4 i.e. greater than 0 for x=-2
therefore its minima

hence f(x) has a minima for x=-2 and minimum value is
2*4+8*(-2)+5
-3 is the answer
and has no minimum value

Answer 8

-3