Find a second degree polynomial f(x) (of the form ax2+bx−4) that has a local extrema at (−1/2,−9/2).
let f(x) = ax^2 + bx - 4
if the vertex is (-1/2, -9/2) then its equation could be
y = a(x + 1/2)^2 - 9/2
= a(x^2 + x + 1/4) - 9/2
= ax^2 + ax + (1/4)a - 9/2
matching them up:
ax^2 <---> ax^2 , ax <---> bx , (1/4)a - 9/2 <---> -4
starting with (1/4)a - 9/2 <---> -4
a - 18 = -16
a = 2
ax <---> bx -----> b=2
so your equation is f(x) = 2x^2 + 2x - 4
check by completing the square:
y = 2x^2 + 2x - 4
= 2(x^2 + x + 1/4 - 1/4) - 4
= 2( (x+1/2)^2 - 1/2 - 4
= 2(x+1/2)^2 - 9/2 , so the vertex is (-1/2, 9/2)
yup, I am right
To find a second degree polynomial with a local extremum at (-1/2, -9/2), we need to first find the values of a, b, and c in the general form of the polynomial: f(x) = ax^2 + bx - 4.
In this case, since we know that the polynomial has a local extremum at (-1/2, -9/2), we can use this information to find these values.
1. Start with the information that the local extremum occurs at the x-coordinate of -1/2. This means the x-value of the local extremum is -1/2. We can write this as x = -1/2.
2. Now let's find the corresponding y-value of the local extremum. Substituting x = -1/2 into the polynomial, we have:
f(-1/2) = a(-1/2)^2 + b(-1/2) - 4
-9/2 = (a/4) - (b/2) - 4
-9/2 = (a/4) - (b/2) - 8/2
-9/2 = (a/4) - (b/2) - 8/2
-9/2 = (a/4) - (b/2) - 8/2
-9/2 = (a/4) - (b/2) - 4
-9/2 = (a - 2b - 16)/4
3. Multiply through by 4 to eliminate the fraction:
-18 = a - 2b - 16
4. Simplify:
a - 2b = 2
5. Since we want the polynomial to have a local extremum at (-1/2, -9/2), we also know that the derivative of the polynomial at x = -1/2 is equal to 0. So, let's find the derivative of the polynomial and set it equal to 0.
f'(x) = 2ax + b
Setting x = -1/2, we have:
f'(-1/2) = 2a(-1/2) + b
0 = -a + b
6. Now we have two equations:
a - 2b = 2 (from step 4)
-a + b = 0 (from step 5)
7. Solve the system of equations obtained in steps 4 and 5.
Adding the equations together, we get:
(-2b + b) + (a - a) = 2 + 0
-b = 2
Solving for b, we find:
b = -2
Substituting b back into one of the equations, we find:
-a + (-2) = 0
-a = 2
a = -2
8. We have found the values of a and b. Substituting these values back into the general form of the polynomial, we get:
f(x) = -2x^2 - 2x - 4
Thus, the second degree polynomial that has a local extremum at (-1/2, -9/2) is f(x) = -2x^2 - 2x - 4.
To find a second degree polynomial with a local extremum at (-1/2, -9/2), we can use the vertex form of a quadratic function. The vertex form is given by:
f(x) = a(x - h)^2 + k
Where (h, k) is the coordinates of the vertex. In this case, the vertex is (-1/2, -9/2), so we have:
f(x) = a(x + 1/2)^2 - 9/2
Now, we need to find the value of "a" to determine the shape of the parabola. Since the extremum is a local minimum (since it goes downward), we need "a" to be positive.
Next, we need to find the value of "a" by using another point on the graph of the polynomial. For simplicity, let's use the y-intercept point (0, -4). Plugging these values into the equation, we get:
-4 = a(0 + 1/2)^2 - 9/2
-4 = (1/4)a - 9/2
-4 + 9/2 = (1/4)a
-8/2 + 9/2 = (1/4)a
1/2 = (1/4)a
a = 2
Now, substituting the value of "a" into the equation, we have:
f(x) = 2(x + 1/2)^2 - 9/2
So, the second degree polynomial that satisfies the given conditions is f(x) = 2(x + 1/2)^2 - 9/2.