Differentiate $y=a(x+b)^2 -8$, then find $a$ and $b$ if the parabola:

Question

a) passes through the origin with the gradient $16$

b) has tangent $y=2x$ at the point $P =(4,8)$

for a) $y'= 2a(x+b)$ then $16=2a(0 +b)$

I'm not sure what to do after this or if it's even right?

Answer 1

Notice, we have $$y=a(x+b)^2-8$$ $$\frac{dy}{dx}=2a(x+b)$$

a) Since, the parabola: $y=a(x+b)^2-8$ passes through the origin hence, substituting the coordinates of $(0, 0)$ in the equation, we get $$0=a(0+b)^2-8\iff ab^2=8 $$ $$ ab=\frac{8}{b}\tag 1$$

Now, gradient at the origin $(0, 0)$ is $16$, hence we have
$$ \frac{dy}{dx}=2a(0+b)=2ab$$$$\iff 2ab=16\iff ab=8\tag 2$$ Now, equating (1) & (2), we get $$\frac{8}{b}=8\iff b=1$$ $$\implies a=\frac{8}{a}=\frac{8}{1}=8$$ Hence, we have $$\bbox[5px, border:2px solid #C0A000]{\color{red}{a=8,\ b=1}}$$

b) Since, the parabola: $y=a(x+b)^2-8$ passes through the point $(4, 8)$ hence, substituting the coordinates of $(4, 8)$ in the equation of parabola, we get $$8=a(4+b)^2-8\iff a(4+b)^2=16 \tag 3$$

Now, gradient at the origin $(4, 8)$ will be equal to the slope of line: $y=2x$
$$\implies \frac{dy}{dx}=2a(4+b)=8a+2ab$$$$\iff 8a+2ab=2$$$$\iff a(4+b)=1$$ $$\iff a=\frac{1}{4+b}\tag 4$$
From (3) & (4), we get $$\frac{1}{4+b}(4+b)^2=16 $$$$\iff 4+b=16\iff b=16-4=12$$

Now, $$a=\frac{1}{4+b}=\frac{1}{4+12}=\frac{1}{16}$$ Hence, $$\bbox[5px, border:2px solid #C0A000]{\color{red}{a=\frac{1}{16},\ b=12}}$$

Answer 2

since parabola go pass true the origin y=a(x+b)^2 -8 must be satisfy (0,0)

there fore a.b^2=8

since you have that a.b =8 ,b must be equal to 1 , and a =8

and lets check y= 8(x+1)^2−8 has a tangent y=2x at(4,8) of not

you have that y'= 2a(x+b) there fore y'= 16(x+1) not equal to 2x

thus there is no parabola which satisfying both conditions