We are given the following:
- parabola passes to both (1,0) and (0,1)
<span> - slope at x = 1 is 4 from the equation of the tangent line </span>
<span>First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. W</span><span>hen x = 0, y = 1. So, c should be equal to 1. The</span><span> parabola is y = ax^2 + bx + 1 </span>
<span>Now, we can substitute the point (1,0) into the equation,
</span>0 = a(1)^2 + b(1) + 1
<span>0 = a + b + 1
a + b = -1 </span>
<span>The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.</span>
<span>We take the derivative of the equation ,
y = ax^2 + bx + 1</span>
<span>y' = 2ax + b
</span>
<span>x = 1, y' = 2
</span>4 = 2a(1) + b
<span>4 = 2a + b </span>
So, we have two equations and two unknowns,<span> </span>
<span>2a + b = 4 </span>
<span>a + b = -1
</span><span>
Solving simultaneously,
a = 5 </span>
<span>b = -6</span>
<span>Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .</span>
Y+4= x blue bikes
G-2= x yellow bikes
4y+4= 8 blue bikes
4y+2= 6 green bikes
4y= 4 yellow bikes
8 blue bikes
+
6 green bikes
+
4 yellow bikes
=
18 bikes total
Ms Mess gave 2*100=200$
The Clerk gave 3*50=150$
200-150=50$
The end up paying 50-70= (-20$)
Not sure
Answer:
a and b are relative on maxima interval x =-4,x=0
