The graph of y = kx-k touch the graph of y = 3x²-4x + 1 at P.Find the coordinates of P.

To find the coordinates of point \( P \) where the graph of \( y = kx - k \) touches the graph of \( y = 3x^2 - 4x + 1 \), follow these steps:

### Step 1: Set the equations equal
At point \( P \), the two graphs must share the same \( x \) and \( y \) coordinates. This means their equations are equal at point \( P \).

Set the two equations equal to each other:

\[
kx - k = 3x^2 - 4x + 1
\]

Rearranging the terms:

\[
0 = 3x^2 - (4 + k)x + (1 + k)
\]

This is a quadratic equation in terms of \( x \).

### Step 2: Apply the "tangent" condition
Since the graphs touch at \( P \), they must have the same slope at \( P \), meaning they have a common point and common tangent. The discriminant of the quadratic equation must be zero for a single solution (since the graphs only touch at one point).

The discriminant \( \Delta \) for a quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[
\Delta = b^2 - 4ac
\]

For our equation \( 3x^2 - (4 + k)x + (1 + k) = 0 \), the coefficients are:
- \( a = 3 \)
- \( b = -(4 + k) \)
- \( c = (1 + k) \)

Now, set the discriminant equal to zero:

\[
(4 + k)^2 - 4(3)(1 + k) = 0
\]

### Step 3: Solve the discriminant equation
Expand both sides:

\[
(4 + k)^2 = 16 + 8k + k^2
\]
\[
4(3)(1 + k) = 12 + 12k
\]

Now substitute these into the discriminant equation:

\[
16 + 8k + k^2 - (12 + 12k) = 0
\]

Simplifying:

\[
k^2 - 4k + 4 = 0
\]

This is a perfect square trinomial:

\[
(k - 2)^2 = 0
\]

Thus, \( k = 2 \).

### Step 4: Find the coordinates of \( P \)
Now that we know \( k = 2 \), substitute this value back into the equation of the line:

\[
y = 2x - 2
\]

To find the point \( P \), substitute \( k = 2 \) into the quadratic equation \( 3x^2 - 4x + 1 \):

\[
2x - 2 = 3x^2 - 4x + 1
\]

Rearranging terms:

\[
0 = 3x^2 - 6x + 3
\]

Dividing through by 3:

\[
x^2 - 2x + 1 = 0
\]

This is also a perfect square trinomial:

\[
(x - 1)^2 = 0
\]

Thus, \( x = 1 \).

Finally, substitute \( x = 1 \) into either equation to find \( y \):

\[
y = 2(1) - 2 = 0
\]

Thus, the coordinates of point \( P \) are \( (1, 0) \).

### Final Answer:
The coordinates of point \( P \) are \( \boxed{(1, 0)} \).

Would you like more details or have any questions?

Here are 5 related questions you might explore:
1. What does it mean for two curves to be tangent to each other?
2. How do you determine the discriminant of a quadratic equation?
3. Can a line intersect a parabola at more than one point? Under what conditions?
4. How do you solve a quadratic equation using the discriminant?
5. How do the slopes of a line and a curve compare at a point of tangency?

**Tip:** Always check the discriminant to determine the number of intersection points when solving systems involving quadratics.

The graph of y = kx - k touches the graph of y = 3x² - 4x + 1 at P. Find the coordinates of P.

Discover how to find the coordinates of point P where the line y = kx - k touches the parabola y = 3x² - 4x + 1. This step-by-step solution covers key algebraic concepts, including solving quadratic equations, applying the discriminant, and using the tangency condition. Perfect for students in Grades 9-11.

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