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For each function, evaluate (a) \(f(0,0)\); (b) \(f(1,0) ;\) (c) \(f(0,-1)\); (d) \(f(a, 2) ;\) (e) \(f(y, x)\);(f) \(f(x+h, y+k)\) HINT [See Quick Examples page 1080.] $$ f(x, y)=x^{2}+y^{2}-x+1 $$

Short Answer

Expert verified
The short answers for the given function, \(f(x,y) = x^2 + y^2 - x + 1\), evaluated at different inputs are: (a) \(f(0,0) = 1\) (b) \(f(1,0) = 1\) (c) \(f(0,-1) = 2\) (d) \(f(a,2) = a^2 - a + 5\) (e) \(f(y,x) = y^2 + x^2 - y + 1\) (f) \(f(x+h, y+k) = x^2 + 2xh + h^2 + y^2 + 2yk + k^2 - x - h + 1\)
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Step 1: (a)

: To find \(f(0,0)\), we plug in \(x = 0\) and \(y = 0\) into the function: $$ f(0, 0) = (0)^2 + (0)^2 - (0) + 1 = 0 + 0 + 1 = 1 $$

Step 2: (b)

: To find \(f(1,0)\), we plug in \(x = 1\) and \(y = 0\) into the function: $$ f(1, 0) = (1)^2 + (0)^2 - (1) + 1 = 1 + 0 - 1 + 1 = 1 $$

Step 3: (c)

: To find \(f(0,-1)\), we plug in \(x = 0\) and \(y = -1\) into the function: $$ f(0, -1) = (0)^2 + (-1)^2 - (0) + 1 = 0 + 1 + 1 = 2 $$

Step 4: (d)

: To find \(f(a, 2)\), we plug in \(x = a\) and \(y = 2\) into the function: $$ f(a, 2) = a^2 + (2)^2 - a + 1 = a^2 + 4 - a + 1 = a^2 - a + 5 $$

Step 5: (e)

: To find \(f(y, x)\), we plug in \(x = y\) and \(y = x\) into the function: $$ f(y, x) = y^2 + x^2 - y + 1 $$

Step 6: (f)

: To find \(f(x+h, y+k)\), we plug in \(x = x + h\) and \(y = y + k\) into the function: $$ f(x+h, y+k) = (x + h)^2 + (y + k)^2 - (x + h) + 1 = x^2 + 2xh + h^2 + y^2 + 2yk + k^2 - x - h + 1 $$

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