graph of f(x)=(x-2)^2-3
using slope of -4
write the equations of the line that intersects the parabola at one point, two points and no points.

2. Originally Posted by banana7
using slope of -4
write the equations of the line that intersects the parabola at one point,
find the eqn of the tangent

$y - f(x_0) = f'(x_0)(x-x_0)$

where $x_0$ is the solution to $f'(x) = -4$

3. Accidental double post.

4. To help in understanding the problem, you might graph the function. It is a parabola, opening upward, with vertex at (2, -3). Then draw several lines with slope "4". Some of them will cross through the parabola, intersecting it in two places. Others will go below the vertex and not intersect at all. Exactly one line will be tangent to it, at a point where the derivative of the quadratic function is 4.

Another way to do this problem, without using Calculus, is this. We can write any line with slope 4 as y= 4x+ b for some number b. That line will intersect the graph of $y= (x- 2)^2- 3$ where $y= (x-2)^2- 3= 4x+ b$. That is a quadratic equation and will have one, two, or zero solutions depending upon whether its "discriminant" is 0, positive, or negative respectively.

(The discriminant of the quadratic $ax^2+ bx+ c$ is $b^2- 4ac$.)