# Math Help - Equation of tangent to parabola

1. ## Equation of tangent to parabola

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 .

2. Originally Posted by goguy123
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 $f'(1)$

In this case $f'(1) = -1$

Now you know the gradient use the straight line equation $y=mx+c$ to find $c$ using the ordered pair $(1,6)$

In this case $c = 7$

So the equation is $y = -x + 7$ or $y= 7-x$

3. Originally Posted by goguy123
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 .
The slope of the line tangent to $f(x)$ at the point $(1.6)$ is $m=f'(1)$.

So, recall that the equation to a line is $(y_2-y_1)=m(x_2-x)$

4. Thanks a lot for the help.

5. Originally Posted by goguy123
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 .
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 $y - 6 = m(x - 1) \Rightarrow y = mx - m + 6$ where m is the gradient.

Also, the tangent (by definition) only intersects the parabola once. So the equations:

$y = 6 + x - x^2$ .... (1)

$y = mx - m + 6$ .... (2)

have only one solution. Solving (1) and (2) simultaneously:

$6 + x - x^2 = mx - m + 6 \Rightarrow x^2 + (m - 1)x - m = 0$.

Since there's only one solution, the discriminant is equal to zero:

$(m - 1)^2 + 4m = 0$.

Solve this equation for m and substitute the value into $y = mx - m + 6$ to get the final equation of the tangent.