<p>We just started derivatives in pre-calc…I don’t understand these two problems from homework.</p>
<li><p>Find a second-degree polynomial P such that P(2) = 5, 1st derivative of 2 = 3, and 2nd derivative of 2 = 2.</p></li>
<li><p>Find a parabola with equation y = ax^2 + bx whose tangent line at (1,1) has equation y = 3x-2.</p></li>
</ol>
<p>I googled the first question and one site said P(x) = 3-x+x^2 and another site said P(x) = x^2-2x+2. Are either of these right? I need to know how to get the answers also!</p>
<p>So you have a second degree polynomial. Let's put it in a general form:
f(x)=Ax^2+Bx+C
Then f'(x)=2Ax+B and f''(x)=2A
But we know f''(2)=2=2A
So A=1.
This then gives us f'(x)=2x+B
f'(2)=4+B=3
So B=-1
Plugging in again, we get:
f(x)=x^2-x+C
f(2)=4-2+C=5
So C=3, and your function is f(x)=x^2-x+3</p>
<ol>
<li>If it's a 2nd degree polynomial, then it's of the form
P(x) = (a)x^2 + bx + c
The first derivative is P'(x) = 2ax + b
and the 2nd derivative is P''(x) = 2a</li>
</ol>
<p>Since P(2) = 5, we have
a(2)^2 + b(2) + c = 5 or 4a + 2b + c = 5 --------------(1)</p>
<p>Since P'(2) = 3, we have
2a(2) + b = 3 or 4a + b = 3 --------------(2)</p>
<p>Since P''(2) = 2, we have
2a = 2 or a = 1 --------------(3)</p>
<p>Solve equations (1),(2) and (3) to get
a=1, b=-1, c=3
So P(x) = ax^2 + bx + c = x^2 - x + 3</p>
<ol>
<li>The tangent line to the parabola at any point has a slope = y'(x)
= 2ax + b</li>
</ol>
<p>At (1,1), y'(1) = 2a + b = slope of straight line 3x -2 = 3
so 2a + b = 3 --------------------(1)</p>
<p>Since the parabola passes through (1,1),
y = ax^2 + b becomes 1 = a(1) + b or
a + b = 1 --------------------(2)
Solve (1) and (2) to get a=2, b=-1</p>