Question

what point on the line y = 7x + 8 is closest to the origin?
the point on the line y = 7x + 8 closest to the origin is
(type an ordered pair using integers or fractions. simplify your answer.)

Explanation:

Step1: Use the distance formula

The distance \(d\) from a point \((x,y)\) to the origin \((0,0)\) is \(d=\sqrt{(x - 0)^2+(y - 0)^2}=\sqrt{x^{2}+y^{2}}\). Since \(y = 7x+8\), we substitute \(y\) into the distance formula: \(d=\sqrt{x^{2}+(7x + 8)^{2}}=\sqrt{x^{2}+49x^{2}+112x + 64}=\sqrt{50x^{2}+112x + 64}\). To simplify, we can minimize the function \(f(x)=50x^{2}+112x + 64\) (because minimizing \(f(x)\) is equivalent to minimizing \(d\) as \(d=\sqrt{f(x)}\) and the square - root function is a monotonic increasing function for non - negative values).

Step2: Find the derivative of \(f(x)\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(f^\prime(x)=(50x^{2}+112x + 64)^\prime=100x+112\).

Step3: Set the derivative equal to zero

Set \(f^\prime(x)=0\), so \(100x+112 = 0\). Solving for \(x\):

$$
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Step4: Find the value of \(y\)

Substitute \(x =-\frac{28}{25}\) into \(y = 7x+8\).

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Answer:

\((-\frac{28}{25},\frac{4}{25})\)