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Find the equation of the tangent to the curve $y = x^2 -6x +5$ at each point where the curve cuts the axis. Find also the coordinates of the point where these tangent line meet. I found the gradient function to be $2x-6$ and I know the curve cuts the x axis when $y =0$

Harry Peter
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Tesha Caesar
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  • substitute $y=0$ in the equation $y=x^2-6x+5$ & solve quadratic equation $x^2-6x+5=0$ or $(x-1)(x-5)=0$ that gives $x=1, 5$ so the points on the curve where it cuts x-axis are $(1, 0)$ & $(5, 0)$ now, find the slopes at these points & find the equations of the tangents. Then solve both the equation to find the intersection point of the tangents. Also see the answer of your previous question http://math.stackexchange.com/a/1576716/210295 – Harish Chandra Rajpoot Dec 15 '15 at 14:14
  • $y-(x-5)(x-1)$ at $(5,0)$ and $(1,0)$ are $y-(x+5-5)(x+5-1)$ and $y-(x+1-5)(x+1-1),$ i.e. zero-set of tangent cone translated back $y=4(x—5)$ and $y=-4(x-1).$ They intersect at $2y=-16,8x-24=0$ or $(x,y)=(3,-8).$ – Jan-Magnus Økland Oct 14 '23 at 08:49

2 Answers2

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You know that the curve cuts the x-axis when y = 0, so solve for x if y = 0

$x^2-6x+5=0$
$x=\frac{6+\sqrt{6^2-4(5)}}{2}, x=\frac{6-\sqrt{6^2-4(5)}}{2}$
$x=1, x=5$

The tangent to the curve should have the same slope as the derivative at x. Also, it should pass through a particular point $(x, f(x))$.

Let's find the derivative of your original function.
$f(x)=x^2-6x+5$
$f'(x)=2x-6$

Now, the slope at $x=1$ is $-4$, and at $x=5$ is $4$

How can we find a general function such that it's slope is $s$ and it passes through $(x_1, y_1)$? Let's let the function be $g(x)$.

If $g'(x) = s$, it follows that $\int g'(x) dx = \int s dx$
$g(x)=s\int{dx}+C$
$g(x) = sx + C$
And of course, on differentiating this, we get the slope to be $s$.

For the second condition, consider $x = x_1$.
$g(x_1)=sx_1+C$
$y_1=sx_1+C$
$C=y_1-sx_1$
$g(x)=sx+y_1-sx_1$
$g(x)=s(x-x_1)+y_1$

Now, all that's left to do is plug in $x_1$ and $y_1$

At $x=1, y=0$, so the equation is $-4x+4$
At $x=5, y=0$, so the equation is $4x-20$

To find the intersection is akin to asking the question: "at what point will x and y be equal to each other?". This can be found by solving these equations simultaneously.

$y = -4x + 4$
$y = 4x - 20$
$4x - 20 = -4x + 4$
$8x = 24$
$x = 3$

$y = 4(3) - 20$
$y = -8$

Therefore, these tangents intersect at $(3, -8)$.

avighnac
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Start by factoring the polynomial. $y=x^2-6x+5=(x-1)(x-5)$. Thus, the graph intercepts the axis at $x=1$ and $x=5$.

The tangent line is given by $y-y_o=m(x-x_o)$.

For $x=1$: The tangent line is given by $y-0=-4(x-1)=-4x+4$

For $x=5$: The tangent line is given by $y-0=19(x-5)=19x-95$.

To find the intersection coordinates, set the two equations equal to each other:

$-4x+4=19x-95$

$99=23x$

$x= \frac{99}{23}$

$y(\frac{99}{23})=(\frac{99}{23})^2-6(\frac{99}{23})+5= \frac{-1216}{529}$

Thus, the intersection coordinate of the tangent line is $(99/23, -1216/529)$

zz20s
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