Respuesta :
Answer:
[tex]y=6x^2-4x-14[/tex]
Step-by-step explanation:
[tex]y=ax^2+bx+c[/tex]
We are given the slopes (the derivatives) at [tex]x=1[/tex] and [tex]x=-1[/tex].
So let's find the derivative of our equation:
[tex]y'=2ax+b[/tex]
At [tex]x=1[/tex] we have [tex]y'=8[/tex]:
[tex]8=2a(1)+b[/tex]
[tex]8=2a+b[/tex]
At [tex]x=-1[/tex] we have [tex]y'=-16[/tex]:
[tex]-16=2a(-1)+b[/tex]
[tex]-16=-2a+b[/tex]
So we have the system:
[tex]8=2a+b[/tex]
[tex]-16=-2a+b[/tex]
We can solve this by elimination. I'm going to add the equations allowing me to solve for [tex]b[/tex].
[tex]-8=2b[/tex]
Divide both sides by 2:
[tex]-4=b[/tex]
Let's find [tex]a[/tex].
[tex]8=2a+b[/tex] with [tex]b=-4[/tex]:
[tex]8=2a+(-4)[/tex]
[tex]8=2a-4[/tex]
Add 4 on both sides:
[tex]12=2a[/tex]
Divide both sides by 2:
[tex]6=a[/tex]
So we have the equation so far is:
[tex]y=6x^2-4x+c[/tex].
We can find [tex]c[/tex] by using the point (2,2):
[tex]2=6(2)^2-4(2)+c[/tex]
[tex]2=6(4)-4(2)+c[/tex]
[tex]2=24-8+c[/tex]
[tex]2=16+c[/tex]
Subtract 16 on both sides:
[tex]-14=c[/tex]
So the equation is [tex]y=6x^2-4x-14[/tex].
Let's do a final check.
[tex]y'=12x-4[/tex]
Check to see if we have (1,8) and (-1,-16) satisfies this. (Note: [tex](x,y')[/tex])
Check for [tex](x,y')=(1,8)[/tex]:
[tex]8=12(1)-4[/tex]
[tex]8=8[/tex] which is true.
Check for [tex](x,y')=(-1,-16)[/tex]:
[tex]y'=12x-4[/tex]
[tex]-16=12(-1)-4[/tex]
[tex]-16=-16[/tex] which is true.
Let's check to see (2,2) is on [tex]y=6x^2-4x-14[/tex]:
[tex]2=6(2)^2-4(2)-14[/tex]
[tex]2=6(4)-8-14[/tex]
[tex]2=24-8-14[/tex]
[tex]2=24-22[/tex]
[tex]2=2[/tex] which is true.
The equation [tex]y=6x^2-4x-14[/tex] has been verified with the given conditions.
