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Find the points on the curve y=x2+5 closest to the point (0,9). Round to the nearest two decimal places. Write the point with the smaller x value first.

Find the points on the curve y=x2+5 closest to the point (0,9). Round to the nearest-example-1

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The equation for the distance between any point and the point (0, 9) is:


d=√(x^2+(y-9)^2)

The curve given in the problem is:


y=x^2+5

If we solve for 'x²', we can substitute in the distance equation and obtain a function for the distance of any point in the curve to (0, 9):


x^2=y-5

And substitute:


d(y)=√((y-5)+(x-9)^2)

SImplify:


d(y)=√(y^2-17y+76)

And now, we need to calculate the first and second derivatives:


d^(\prime)(x)=(2y-17)/(2√(y^2-17y+76))
d^(\prime)^(\prime)(y)=\frac{15}{4(y^2-17y+76)^{(3)/(2)}}

Then, find the critical points of d(y). Since this function is a quotient, it will be equal to 0 when the numerator is equal to 0:


\begin{gathered} 2y-17=0 \\ . \\ y=(17)/(2) \end{gathered}

Now, evaluate this value into the second derivative:


d^(\prime)^(\prime)((17)/(2))=\frac{15}{4(((17)/(2))^2-17\cdot(17)/(2)+76)^{(3)/(2)}}\approx0.516546

Since is a positive value, the function d(y) has a minimum in y = 17/2

Next, we need to find the values of x. We use the equation of the curve:


\begin{gathered} (17)/(2)=x^2+5 \\ . \\ x^2=(17)/(2)-5 \\ . \\ x=\pm\sqrt{(7)/(2)}=\pm(√(14))/(2) \end{gathered}

Thus, the point on the curve closest to (0, 9), rounded to two decimals, are:


(-1.87,8.5)
(1.87,8.5)

User Tanysha
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