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Sagot :
The locations on the function's graph that are most near the specified point f(x) = x2, (0,9) are approximately 3 units to the right and left, respectively, or (sqr8.5, 8.5).
The initial estimate would be rough to the right and left by around 3 units, or (sqr8.5, 8.5), rounded to the nearest decimal place.
Visually, it is most likely just below y=9 and just below x=3. Additionally, roughly (-3, 9) slightly lower than 3 and 9
Connect the parabola's nearest point to the point at (0,9). Where the two intersect, its slope is the negative inverse of the parabola's slope. The derivative's slope is 2x for the equation f(x) = x2. -1/2x is its inverse negative. The line from (0,9) to the parabola's nearest point has a slope that is perpendicular to the tangent to the parabola at that location.
Get the equation of the line joining (0,9) and the nearest point (x,y) on the parabola using the point-slope formula: 8.5 or y=-1/2 + 9 = (-1/2x)(x)
The closest point's y coordinate is that. The square root of 8.5, including both positive and negative square roots, is the closest point's x coordinate (-sqr8.5, 8.5)
Sqr8.5 =[tex]\sqrt{8.5}[/tex], or approximately 2.91 or -2.91
The process entails finding the equation for the line connecting (0,9) with the parabola's nearest point and then setting the two equations equal. Set the line equation to y=(-1/2x), the parabola equation. The parabola equation is y=[tex]x^{2}[/tex] and the line equation is x + 9.
Equalize them, then find x using the square of the result. The closest point is at x,y. There are two places that are equally far from (0,9), both of which are as close as they can be to 3 at (3,9). (0,9). At the parabola point, that line is not perpendicular to the tangent line (3,9). The line must be perpendicular to the tangent line, which is the slope of the parabola, in order to be closest.
To learn more about parabolas, refer:-
https://brainly.com/question/4074088
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