Question:
The curve
\[
y=ax^3+bx^2+cx+5
\]
touches the \(x\)-axis at \(P(-2,0)\), then \(c=\)
The curve
\[
y=ax^3+bx^2+cx+5
\]
touches the \(x\)-axis at \(P(-2,0)\), then \(c=\)
Show Hint
If a curve touches the \(x\)-axis at a point, then the point satisfies the curve equation and the derivative at that point is zero.
Updated On: Jun 26, 2026
- \(4a+5\)
- \(4a-5\)
- \(5-4a\)
- \(0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the condition that the curve passes through \(P(-2,0)\).
Since the curve touches the \(x\)-axis at \[ P(-2,0), \] the point lies on the curve. Therefore, \[ 0=a(-2)^3+b(-2)^2+c(-2)+5. \] So, \[ 0=-8a+4b-2c+5. \] Hence, \[ -8a+4b-2c+5=0. \]
Step 2: Use the touching condition.
Since the curve touches the \(x\)-axis at \(P(-2,0)\), the tangent at this point is the \(x\)-axis. Therefore, \[ \frac{dy}{dx}=0 \] at \[ x=-2. \] Now, \[ y=ax^3+bx^2+cx+5. \] Differentiating, \[ \frac{dy}{dx}=3ax^2+2bx+c. \] At \(x=-2\), \[ 0=3a(-2)^2+2b(-2)+c. \] So, \[ 0=12a-4b+c. \] Hence, \[ c=4b-12a. \]
Step 3: Substitute \(c=4b-12a\) in the first equation.
From Step 1, \[ -8a+4b-2c+5=0. \] Substitute \[ c=4b-12a. \] Then, \[ -8a+4b-2(4b-12a)+5=0. \] Simplifying, \[ -8a+4b-8b+24a+5=0. \] Thus, \[ 16a-4b+5=0. \] So, \[ 4b=16a+5. \] Hence, \[ b=4a+\frac54. \]
Step 4: Find \(c\).
Using \[ c=4b-12a, \] we get \[ c=4\left(4a+\frac54\right)-12a. \] Therefore, \[ c=16a+5-12a. \] Hence, \[ c=4a+5. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{4a+5}. \]
Since the curve touches the \(x\)-axis at \[ P(-2,0), \] the point lies on the curve. Therefore, \[ 0=a(-2)^3+b(-2)^2+c(-2)+5. \] So, \[ 0=-8a+4b-2c+5. \] Hence, \[ -8a+4b-2c+5=0. \]
Step 2: Use the touching condition.
Since the curve touches the \(x\)-axis at \(P(-2,0)\), the tangent at this point is the \(x\)-axis. Therefore, \[ \frac{dy}{dx}=0 \] at \[ x=-2. \] Now, \[ y=ax^3+bx^2+cx+5. \] Differentiating, \[ \frac{dy}{dx}=3ax^2+2bx+c. \] At \(x=-2\), \[ 0=3a(-2)^2+2b(-2)+c. \] So, \[ 0=12a-4b+c. \] Hence, \[ c=4b-12a. \]
Step 3: Substitute \(c=4b-12a\) in the first equation.
From Step 1, \[ -8a+4b-2c+5=0. \] Substitute \[ c=4b-12a. \] Then, \[ -8a+4b-2(4b-12a)+5=0. \] Simplifying, \[ -8a+4b-8b+24a+5=0. \] Thus, \[ 16a-4b+5=0. \] So, \[ 4b=16a+5. \] Hence, \[ b=4a+\frac54. \]
Step 4: Find \(c\).
Using \[ c=4b-12a, \] we get \[ c=4\left(4a+\frac54\right)-12a. \] Therefore, \[ c=16a+5-12a. \] Hence, \[ c=4a+5. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{4a+5}. \]
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