Graphing Tangents to a Quadratic Curve

[pooker]

Homework Statement

Find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together

Homework Equations

y=3-x^2 (-1,2)

The Attempt at a Solution

slope =

f(x + h) + f(x) / h

f(-1 + h) + 2 / h

3- (-x + h)^2 + 2 / h

3 - 1 + 2h + H^2 + 2 / h

2h + h^2 / h

= 2 + h
= 2

slope is 2

y - y1 = m(x - x1)

y - 2 = 2 (x + 1)
y = 2x + 4 = equation Ok so I have no idea how to graph a tangent, can someone tell me how you graph this? I know how to sketch the graph 3- x^2 since it is a ^2 function it is u shaped, flipped on the x-axis moved up three on the y. I know the tangent obviously passes through -1, 2 but how do I graph the rest? I know y intercept for tangent is 4, but where does it pass through x at? I am a little rusty

 

[Dick]

Do you mean where does it cross the x-axis? It does that when y=0 in the equation y=2x+4. Solve for x. You are rusty. It's just graphing a straight line.