Got a Prelim on monday , please help, thanks
Q: Find the equation of the tangent to the parabola with equation
y=6+x-x(squared) at the point (1,6 ) .
If possible please explain to me how to do this question thanks .
Got a Prelim on monday , please help, thanks
Q: Find the equation of the tangent to the parabola with equation
y=6+x-x(squared) at the point (1,6 ) .
If possible please explain to me how to do this question thanks .
If you've done (or supposed to have done) calculus take the derivative and find $\displaystyle f'(1)$
In this case $\displaystyle f'(1) = -1$
Now you know the gradient use the straight line equation $\displaystyle y=mx+c$ to find $\displaystyle c$ using the ordered pair $\displaystyle (1,6)$
In this case $\displaystyle c = 7$
So the equation is $\displaystyle y = -x + 7$ or $\displaystyle y= 7-x$
Since you did not post this question in the Calculus subforum I assume a non-calculus approach is required ....?
Since the point (1, 6) lies on the tangent line, the equation of the tangent is $\displaystyle y - 6 = m(x - 1) \Rightarrow y = mx - m + 6$ where m is the gradient.
Also, the tangent (by definition) only intersects the parabola once. So the equations:
$\displaystyle y = 6 + x - x^2$ .... (1)
$\displaystyle y = mx - m + 6$ .... (2)
have only one solution. Solving (1) and (2) simultaneously:
$\displaystyle 6 + x - x^2 = mx - m + 6 \Rightarrow x^2 + (m - 1)x - m = 0$.
Since there's only one solution, the discriminant is equal to zero:
$\displaystyle (m - 1)^2 + 4m = 0$.
Solve this equation for m and substitute the value into $\displaystyle y = mx - m + 6$ to get the final equation of the tangent.