To find the point on the graph of y=x^2+1 that's closest to the point (8, 1.5), we first need to find the distance between any point on the graph and the point (8, 1.5).
Let the point on the graph be (x, x^2+1).
The distance between (x, x^2+1) and (8, 1.5) is given by the distance formula:
d = sqrt((x - 8)^2 + (x^2+1 - 1.5)^2)
Simplify the distance formula:
d = sqrt((x - 8)^2 + (x^2 - 0.5)^2)
d = sqrt(x^2 - 16x + 64 + x^4 - x + 0.25)
To find the point on the graph that's closest to the point (8, 1.5), we need to minimize the distance function d. Let's find the derivative of d with respect to x and set it equal to 0 to find the minimum distance.
Taking the derivative of d:
d' = 0.5(2x - 16) + 2x^3 - 2 = 0
Simplify:
x - 8 + 2x^3 - 2 = 0
2x^3 + x - 10 = 0
This is a cubic equation that can be solved numerically to find the value of x. Once we find the value of x, we can find the corresponding y value on the graph of y=x^2+1.
Therefore, the point on the graph of y=x^2+1 that's closest to the point (8, 1.5) can be found using the distance formula and minimizing the distance function.
Find the point on the graph of y=x^2+1 hat’s closest to the point 8, 1.5 . Hint: Remember
the distance formula. The distance between two points in a plane is � 2 d � �x1 x2 �y1 � 2 y2 .
1 answer
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