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(a) Find equations of both lines through the point (2, -3) that are tangent to the parabola $ y = x^2 + x. $(b) Show that there is no line through the point (2,7) that is tangent to the parabola. Then draw a diagram to see why.

$y=11(x-2)+3, y=-(x-2)+3$, the reason we cannot find a tangent to the curve from the point $(2,7)$ is because $f''(x)>0$ for all $x$. Where a function is concave up, the tangent lines must fall below the curve. This point is above the curve, and therefore will not have a tangent line to the curve.

07:03

Frank L.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 1

Derivatives of Polynomials and Exponential Functions

Derivatives

Differentiation

Missouri State University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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it's clear. So when you right here. So for party, we're gonna find some equations. So the probable a y equals X squared plus X is in the form X comma X square plus Annette's And this is assuming the tension is that a comma a square plus a and passes through to common negative three We know the slope of attention is a derivative So when we differentiate it we got two x plus one So this is gonna be to a plus one This the tangent passes through the slope Um, the points a Kama Square plus eight and two comma three. So the slope iss be square was a minus Negative three over a minus two Just equal to a square less a plus three over a minus two. This could be equated to to a plus one when we get values of negative one and negative fought negative one and positive five for aid. So there's two tangents that passed through to common negative three. So we're gonna find the equation of detention when a is equal to negative one on and we get why minus negative three over X minus two equals negative one. We get X plus y plus one is equal to zero. Then for a is equal to negative, it is equal to positive. Five. We're gonna do the same thing to get. Why is equal to 11 x minus 25 and then for part B. We have the points. Um, attention is that a comma? He's square plus a and passes through two comma. Seven. You differentiate to get two X plus one. This becomes to a plus one, and this is equal to the slope of the tangent, which is a square plus a minus seven over a square. A minus two. We got a square minus for a plus five on DDE, equal to zero. This is ah, negative discriminative. So there's no real solution when we graph it. We could see that when we plot our graph two comma, seven lies above and doesn't touch it. So no tension passes through the point

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