differentiation calculus qn (1 Viewer)

Masaken · Jun 1, 2022

hello, i've found the answer to part a, which i've found is x + y(b-4)^2 = 2b-4. but with part b (for both i and ii), i've subbed in b = 0 and b = 6 into the equations and i've gotten answers completely different to the correct ones (i got x + 16y + 4 = 0 and x + 4y - 8 = 0 respectively). the answers in the textbook say part a is correct, and the worked solutions also say i have to sub in b = 0 and b = 6 yet they're getting different answers to mine. where did i go wrong?

Masaken · Jun 1, 2022

yeah i'm not sure either, the textbook's answers for those two are x + 4y = 0 for part i, and x + y = 6 for part ii ;-;

jimmysmith560 · Jun 1, 2022

Would the following working help?

$\begin{align*}x+y(b-4)^2&=2b-4 \\\textup{Tangent through the origin:} \\\textup{Sub in }x&=0\, ,y=0 \\0+0(b-4)^2&=2b-4 \\2b&=4 \\b&=2 \\\textup{When }b=2 &\textup{ the tangent equation is:} \\x+y(2-4)^2&=2(2)-4 \\x+4y&=0 \\\textup{Tangent through the W(6,0):} \\\textup{Sub in }x&=6\, ,y=0\\6+0(b-4)^2&=2b-4\\2b&=10\\b&=5 \\\textup{When }b=5 &\textup{ the tangent equation is:} \\x+y(5-4)^2&=2(5)-4 \\x+y&=6 \\\end{align*}$

ExtremelyBoredUser · Jun 1, 2022

jimmysmith560 said:
Would the following working help?

$\begin{align*}x+y(b-4)^2&=2b-4 \\\textup{Tangent through the origin:} \\\textup{Sub in }x&=0\, ,y=0 \\0+0(b-4)^2&=2b-4 \\2b&=4 \\b&=2 \\\textup{When }b=2 &\textup{ the tangent equation is:} \\x+y(2-4)^2&=2(2)-4 \\x+4y&=0 \\\textup{Tangent through the W(6,0):} \\\textup{Sub in }x&=6\, ,y=0\\6+0(b-4)^2&=2b-4\\2b&=10\\b&=5 \\\textup{When }b=5 &\textup{ the tangent equation is:} \\x+y(5-4)^2&=2(5)-4 \\x+y&=6 \\\end{align*}$

yea this makes sense with the part a equation

Masaken · Jun 1, 2022

honestly mate

jimmysmith560 said:
Would the following working help?

$\begin{align*}x+y(b-4)^2&=2b-4 \\\textup{Tangent through the origin:} \\\textup{Sub in }x&=0\, ,y=0 \\0+0(b-4)^2&=2b-4 \\2b&=4 \\b&=2 \\\textup{When }b=2 &\textup{ the tangent equation is:} \\x+y(2-4)^2&=2(2)-4 \\x+4y&=0 \\\textup{Tangent through the W(6,0):} \\\textup{Sub in }x&=6\, ,y=0\\6+0(b-4)^2&=2b-4\\2b&=10\\b&=5 \\\textup{When }b=5 &\textup{ the tangent equation is:} \\x+y(5-4)^2&=2(5)-4 \\x+y&=6 \\\end{align*}$

you are everything the worked solutions aren't

thanks!!

differentiation calculus qn (1 Viewer)

Masaken

Unknown Member

Masaken

Unknown Member

jimmysmith560

Le Phénix Trilingue

ExtremelyBoredUser

Bored Uni Student

Masaken

Unknown Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)