determine the quadritic curve y=f(x) if it touches the line y=x at the point x=1 and passes through the point (-1,0).

To determine the quadratic curve \( y = f(x) \) that touches the line \( y = x \) at the point \( x = 1 \) and also passes through the point \((-1, 0)\), we can start by expressing the quadratic function in its standard form. A general quadratic function can be written as:

Setting Up the Quadratic Function

Let’s define the quadratic function as:

\( f(x) = ax^2 + bx + c \)

Here, \( a \), \( b \), and \( c \) are constants that we need to determine.

Condition 1: Touching the Line \( y = x \)

Since the curve touches the line \( y = x \) at \( x = 1 \), this means two things:

• The function value at \( x = 1 \) must equal \( 1 \): \( f(1) = 1 \).
• The slope of the function at this point must equal the slope of the line \( y = x \), which is \( 1 \): \( f'(1) = 1 \).

Calculating \( f(1) = 1 \)

Substituting \( x = 1 \) into the quadratic function gives:

\( f(1) = a(1)^2 + b(1) + c = a + b + c = 1 \)

Calculating \( f'(1) = 1 \)

The derivative of the quadratic function is:

\( f'(x) = 2ax + b \)

Now, substituting \( x = 1 \) into the derivative:

\( f'(1) = 2a(1) + b = 2a + b = 1 \)

Condition 2: Passing Through the Point \((-1, 0)\)

We also know that the curve passes through the point \((-1, 0)\), which gives us another equation:

\( f(-1) = a(-1)^2 + b(-1) + c = a - b + c = 0 \)

System of Equations

Now we have a system of three equations:

• 1. \( a + b + c = 1 \)
• 2. \( 2a + b = 1 \)
• 3. \( a - b + c = 0 \)

Solving the System

We can solve this system step by step. First, let’s express \( c \) from the first equation:

\( c = 1 - a - b \)

Now, substitute \( c \) into the third equation:

\( a - b + (1 - a - b) = 0 \)

Simplifying this gives:

\( 1 - 2b = 0 \Rightarrow b = \frac{1}{2} \)

Next, substitute \( b = \frac{1}{2} \) into the second equation:

\( 2a + \frac{1}{2} = 1 \Rightarrow 2a = \frac{1}{2} \Rightarrow a = \frac{1}{4} \)

Now, substitute \( a \) and \( b \) back into the equation for \( c \):

\( c = 1 - \frac{1}{4} - \frac{1}{2} = 1 - \frac{3}{4} = \frac{1}{4} \)

Final Quadratic Function

Now that we have \( a \), \( b \), and \( c \), we can write the quadratic function:

\( f(x) = \frac{1}{4}x^2 + \frac{1}{2}x + \frac{1}{4} \)

Verification

To ensure our function meets the conditions:

• At \( x = 1 \): \( f(1) = \frac{1}{4}(1)^2 + \frac{1}{2}(1) + \frac{1}{4} = 1 \) (correct).
• At \( x = 1 \): \( f'(1) = 2(\frac{1}{4})(1) + \frac{1}{2} = 1 \) (correct).
• At \( x = -1 \): \( f(-1) = \frac{1}{4}(-1)^2 + \frac{1}{2}(-1) + \frac{1}{4} = 0 \) (correct).

Thus, the quadratic curve that satisfies both conditions is:

\( y = \frac{1}{4}x^2 + \frac{1}{2}x + \frac{1}{4} \)