Question

a tangent to the graph of $y = x^{2}+1$ is parallel to the line $y = x$. find the $y$-coordinate of the point where this tangent touches the graph. give your answer as an integer.

Explanation:

Step1: Find the derivative of the function

The function is \(y = x^{2}-\frac{1}{x}\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\) and \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) (here \(u = 1\), \(v=x\)), the derivative \(y^\prime=2x+\frac{1}{x^{2}}\).

Step2: Set the derivative equal to the slope of the given line

The line \(y = x\) has a slope \(m = 1\). So we set \(2x+\frac{1}{x^{2}}=1\). Multiply through by \(x^{2}\) (assuming \(x
eq0\)) to get \(2x^{3}+1=x^{2}\), or \(2x^{3}-x^{2}+1 = 0\).
We can try some simple integer values. When \(x=-1\), \(2(-1)^{3}-(-1)^{2}+1=2\times(-1)-1 + 1=-2
eq0\). When \(x = 1\), \(2(1)^{3}-(1)^{2}+1=2 - 1+1=2
eq0\).
Let's rewrite the equation as \(2x^{3}-x^{2}+1=(x + 1)(2x^{2}-3x + 1)\). Factoring \(2x^{2}-3x + 1=(2x - 1)(x - 1)\). The roots of \(2x^{3}-x^{2}+1 = 0\) are \(x=-1,\frac{1}{2},1\).
Substitute \(x=\frac{1}{2}\) into \(y = x^{2}-\frac{1}{x}\), \(y=(\frac{1}{2})^{2}-\frac{1}{\frac{1}{2}}=\frac{1}{4}-2=-\frac{7}{4}\) (not an integer).
Substitute \(x = 1\) into \(y = x^{2}-\frac{1}{x}\), \(y=1^{2}-\frac{1}{1}=0\) (an integer).

Answer:

\(0\)