find a function y=ax^2+bx+c whose graph has an x-intercept of 1, a y-intercept of -2, and a tangent line with slope -1 at the y-intercept.

Why did the parabola bring a clown to the party?

Because it wanted to add some humor to the intercepts!

To find a function that meets the given conditions, we can follow these steps:

Step 1: Use the x-intercept to find a factor of the quadratic equation.
Given that the x-intercept is 1, we know that the quadratic equation can be written in the factored form as (x - 1)(x + p), where p is a constant.

Step 2: Use the y-intercept to find the value of c.
Since the y-intercept is -2, the value of c is -2.

Step 3: Use the tangent line slope to find the value of b.
Since the tangent line slope at the y-intercept is -1, the derivative of the quadratic equation evaluated at x = 0 should equal -1. Taking the derivative of the quadratic equation yields 2ax + b, and setting x = 0 gives b = -1.

Step 4: Use the values of a, b, and c to write the equation in standard form.
The equation in standard form is y = ax^2 + bx + c. Plugging in the values we have determined, we get y = ax^2 - x - 2.

Therefore, a function that satisfies the given conditions is y = ax^2 - x - 2, where a is a constant.

To find a function with the given characteristics, we will work through the information step by step.

1. Given that the graph has an x-intercept of 1, we know that when x = 1, y = 0. This gives us our first point on the graph: (1, 0).

2. Given that the graph has a y-intercept of -2, we know that when y = -2, x = 0. This gives us our second point on the graph: (0, -2).

3. We are also given that the tangent line at the y-intercept has a slope of -1. This implies that the derivative of the function at x = 0 is -1. Differentiating the quadratic function y = ax^2 + bx + c will give us its derivative.

Let's go through the steps one by one:

Step 1: Find the derivative of the quadratic function.
The derivative of y = ax^2 + bx + c with respect to x is given by dy/dx = 2ax + b.

Step 2: Use the derivative to find the slope.
Plug in x = 0 into the derivative equation: 2a(0) + b = -1.
This simplifies to b = -1.

Step 3: Use the given points to find the values of a and b.
We already found b = -1, and from the y-intercept (0, -2) we know that the function gives y = -2 when x = 0.
Substituting these values into the quadratic equation: -2 = a(0)^2 + (-1)(0) + c, which simplifies to c = -2.

Step 4: Use the x-intercept point to find a.
From the x-intercept (1, 0), we know that the function gives y = 0 when x = 1.
Substituting these values into the quadratic equation: 0 = a(1)^2 + (-1)(1) + (-2), which simplifies to a - 1 - 2 = 0.
This results in a = 3.

Putting it all together, we have found the values of a, b, and c. The function that satisfies the given conditions is: y = 3x^2 - x - 2.

The three given conditions are:

1. For x-intercept of 1:
f(x)=y= (x-1)(ax-c) = ax²-(c+1)+c
2. For y-intercept of -2:
c=-2
3. for tangent=-1 at y=0
Now find
f'(x)=2ax+b
to satisfy
f'(0)=-1, or
2a(0)+b=-1

Solve for a, b and c.