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To find the parabola with equation \( y = ax^2 + bx \) that satisfies the given conditions, let's break down the information provided:

1. **The point (1,1)** lies on the parabola.
2. **The tangent line at (1,1)** has the equation \( y = 3x - 2 \).

### Step 1: Using the Point (1,1)
Since \( (1,1) \) is a point on the parabola, we can substitute \( x = 1 \) and \( y = 1 \) into the equation:
\[
1 = a(1)^2 + b(1) \Rightarrow 1 = a + b
\]
So, we have the equation:
\[
a + b = 1 \quad \text{(Equation 1)}
\]

### Step 2: Using the Tangent Line Information
The slope of the tangent line at \( x = 1 \) is given by the line \( y = 3x - 2 \), which has a slope of 3. 

Since the derivative \( \frac{dy}{dx} \) of the parabola \( y = ax^2 + bx \) gives the slope of the tangent, we differentiate with respect to \( x \):
\[
\frac{dy}{dx} = 2ax + b
\]
At the point \( x = 1 \), the slope of the tangent (derivative) should equal 3:
\[
2a(1) + b = 3 \Rightarrow 2a + b = 3
\]
This gives us a second equation:
\[
2a + b = 3 \quad \text{(Equation 2)}
\]

### Step 3: Solving the System of Equations
Now, we have two equations:
1. \( a + b = 1 \)
2. \( 2a + b = 3 \)

Subtract Equation 1 from Equation 2:
\[
(2a + b) - (a + b) = 3 - 1
\]
\[
a = 2
\]

Now, substitute \( a = 2 \) into Equation 1:
\[
2 + b = 1
\]
\[
b = -1
\]

### Step 4: Writing the Equation of the Parabola
Now that we have \( a = 2 \) and \( b = -1 \), the equation of the parabola is:
\[
y = 2x^2 - x
\]

### Final Answer
The equation of the parabola is:
\[
\boxed{y = 2x^2 - x}
\]

Would you like further details on any step, or have any questions?

Here are some additional questions that build on this topic:
1. How would you find the vertex of the parabola \( y = 2x^2 - x \)?
2. What is the axis of symmetry for the parabola \( y = 2x^2 - x \)?
3. How can you determine if a point lies inside, on, or outside the parabola?
4. How would the parabola change if the coefficient \( a \) were negative?
5. How can you use calculus to find the minimum or maximum point on a parabola?

**Tip:** When finding a tangent line to a curve at a specific point, use the derivative to obtain the slope and then apply the point-slope formula to write the equation of the line.

Find the parabola with equation y = ax^2 + bx whose tangent line at (1, 1) has equation y = 3x - 2.

Learn how to find the parabola y = ax^2 + bx whose tangent at the point (1,1) has the equation y = 3x - 2. This problem involves quadratic equations, derivatives, and finding the slope of a tangent line, suitable for Grades 11-12 students.

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