Question

find the parabola with equation y = ax^2 + bx whose tangent line at (1, 3) has equation y = 7x - 4.

1. -/2 points

for what values of a and b is the line 4x + y = b tangent to the parabola y = ax^2 when x = 3?

Explanation:

Step1: Use the fact that the point lies on the parabola

Since the point $(1,3)$ lies on the parabola $y = ax^{2}+bx$, substitute $x = 1$ and $y = 3$ into the equation:
$3=a\times1^{2}+b\times1$, so $a + b=3$, which can be rewritten as $b = 3 - a$.

Step2: Find the derivative of the parabola

Differentiate $y=ax^{2}+bx$ with respect to $x$ using the power - rule. $y^\prime=2ax + b$.

Step3: Use the slope of the tangent line

The slope of the tangent line $y = 7x-4$ is $7$. At $x = 1$, the slope of the tangent to the parabola is equal to the slope of the given tangent line. So, when $x = 1$, $y^\prime=2a\times1 + b=7$, which gives $2a + b=7$.

Step4: Substitute $b = 3 - a$ into $2a + b=7$

Substitute $b = 3 - a$ into $2a + b=7$:
$2a+(3 - a)=7$.
Simplify the left - hand side: $2a+3 - a=a + 3$.
So, $a+3 = 7$, and $a = 4$.

Step5: Find the value of $b$

Substitute $a = 4$ into $b = 3 - a$, we get $b=3 - 4=-1$.
The equation of the parabola is $y = 4x^{2}-x$.

Step1: Find the derivative of the parabola

Differentiate $y = ax^{2}$ with respect to $x$ using the power - rule. $y^\prime=2ax$.

Step2: Find the slope of the tangent line at $x = 3$

When $x = 3$, the slope of the tangent to the parabola is $y^\prime=2a\times3 = 6a$.
Rewrite the line $4x + y=b$ in slope - intercept form $y=-4x + b$. The slope of this line is $-4$.
Since the slope of the tangent line to the parabola at $x = 3$ is equal to the slope of the given line, we have $6a=-4$, so $a=-\frac{2}{3}$.

Step3: Find the point on the parabola at $x = 3$

When $x = 3$ and $a = -\frac{2}{3}$, $y=a\times3^{2}=-\frac{2}{3}\times9=-6$.

Step4: Find the value of $b$

The point $(3,-6)$ lies on the line $4x + y=b$. Substitute $x = 3$ and $y=-6$ into the line equation:
$4\times3+( - 6)=b$.
$12-6=b$, so $b = 6$.

Answer:

$y = 4x^{2}-x$

Now for the second sub - question: