One number is 10 larger than another. Write a function in standard form to represent the sum of their squares. The minimum sum of squares is ___.
So I said that the function is f(x)=x^2+x+10 and since the question comes from an assignment on completing the square to convert to vertex form, I completed the square and got f(x)=(x+1/2)^2-39/4. But, now I don<t know what to do to find the minimum sum.
Oh, the function should be f(x)=2x^2+100 right?
smaller number --- x
larger number ---- (x+10)
sum of their squares
= (x+10) + x^2
= x^2 + 20x + 100 + x^2
= 2x^2 + 20x + 100
this is a parabola opening upwards so it has a minimum.
That minimum happens at the vertex.
The x of the vertex is -20/4 = -5
so one number is -5, the other is +5
for a minimum of 25+25 = 50
I'm a little confused: Why isn't x+10 squared?
Oh never mind, I got it! Thank you so much. This helps a lot!! :)
To find the minimum sum of squares of the given function, you can utilize the vertex form of a quadratic function. The vertex form of a quadratic function is:
f(x) = a(x - h)^2 + k
where (h, k) represents the vertex of the parabola.
In this case, the function f(x) can be rewritten as:
f(x) = (x + 1/2)^2 - 39/4
Comparing it with the vertex form, you can see that the vertex is at (-1/2, -39/4). The x-coordinate of the vertex represents the value of x that will result in the minimum value of the function.
To find the minimum sum, we need to find the corresponding y-value of the vertex, which is the minimum value of the function. In this case, the minimum sum of squares is the value of f(x) at the vertex.
So, to find the minimum sum solution, substitute the x-coordinate of the vertex (-1/2) into the function:
f(-1/2) = (-1/2 + 1/2)^2 - 39/4
Simplifying the equation:
f(-1/2) = (0)^2 - 39/4
f(-1/2) = -39/4
Therefore, the minimum sum of squares for the given function is -39/4.