Question

the curve y = ax^2 + bx + c passes through the point (1,9) and is tangent to the line y = 4x at the origin. find a, b, and c.
a = , b = , c =

Explanation:

Step1: Use the fact that the curve passes through (0,0)

Since the curve $y = ax^{2}+bx + c$ is tangent to $y = 4x$ at the origin, the curve passes through the point $(0,0)$. Substitute $x = 0$ and $y=0$ into the equation of the curve:
$0=a(0)^{2}+b(0)+c$, so $c = 0$.

Step2: Use the derivative and the slope of the tangent - line

The derivative of $y=ax^{2}+bx + c$ is $y'=2ax + b$. The slope of the line $y = 4x$ is 4. At the origin ($x = 0$), the slope of the tangent - line to the curve is equal to the value of the derivative at $x = 0$. So, when $x = 0$, $y'(0)=4$. Substitute $x = 0$ into $y'=2ax + b$:
$y'(0)=2a(0)+b$, so $b = 4$.

Step3: Use the fact that the curve passes through (1,9)

Since $c = 0$ and $b = 4$, the equation of the curve is $y=ax^{2}+4x$. The curve passes through the point $(1,9)$. Substitute $x = 1$ and $y = 9$ into the equation:
$9=a(1)^{2}+4(1)$.
$9=a + 4$.
Solve for $a$: $a=9 - 4=5$.

Answer:

$a = 5$, $b = 4$, $c = 0$