Answer:
Step-by-step explanation:
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Answer:
Step-by-step explanation:
Given that a tangent is drawn from P (6,2) to unit circle with center at the origin.
The tangent passing through (6,2) would have equaiton of the form
for suitable m.
Because this line is tangent, distance of centre of circle namely origin is the radius 1.
i.e. 
the coordinates are the intersection of the tangent with the circle
They are (-0.162, 0.987) and (0.462,-0.887)
Is there supposed to be a picture with tr
If
<em>f(x)</em> = <em>ax</em> ³ + <em>bx</em> ² - 5<em>x</em> + 9
then
<em>f '(x)</em> = 3<em>ax </em>² + 2<em>bx</em> - 5
Given that <em>f</em> (-1) = 12 and <em>f</em> '(-1) = 3, we get the system of equations
-<em>a</em> + <em>b</em> + 5 + 9 = 12
3<em>a</em> - 2<em>b</em> - 5 = 3
or
-<em>a</em> + <em>b</em> = -2
3<em>a</em> - 2<em>b</em> = 8
Multiply through the first equation by 2 and add it to the second one to eliminate <em>b</em> and solve for <em>a</em> :
2(-<em>a</em> + <em>b</em>) + (3<em>a</em> - 2<em>b</em>) = 2(-2) + 8
-2<em>a</em> + 2<em>b</em> + 3<em>a</em> - 2<em>b</em> = -4 + 8
<em>a</em> = 4
Substitute this into the first equation above to solve for <em>b</em> :
-4 + <em>b</em> = -2
<em>b</em> = 2
