MHB Where Should I Begin with Parabolas?

[Ilikebugs]

View attachment 6358 I don't know where to start :s

 

[MarkFL]

We know the axis of symmetry for the general quadratic:

$$f(x)=ax^2+bx+c$$

is the line:

$$x=-\frac{b}{2a}$$

And so for the given function:

$$y(x)=5x^2+ax+b$$

The axis of symmetry is:

$$x=-\frac{a}{10}$$

And so the minimum value of $y$ will be:

$$y_{\min}=y\left(-\frac{a}{10}\right)=5\left(-\frac{a}{10}\right)^2+a\left(-\frac{a}{10}\right)+b=\frac{20b-a^2}{20}$$

Now, we are given two points on the parabola, and using this data, we obtain:

$$5a^2+a(a)+b=b$$

$$5b^2+ab+b=a$$

Bearing in mind that $a\ne b$, can you obtain a unique solution to the above system?

 

[Ilikebugs]

a is 0, b is -1/5

 

[MarkFL]

a is 0, b is -1/5

Yes, that's what I got too. (Yes)

So then, what is $y_{\min}$?

 

[Ilikebugs]

-1/5?

 

[Greg]

-1/5?

That's correct.

Alternatively, we have

$$b=5(a)^2+a(a)+b\implies a=0$$

Substituting $-\frac{a}{10}=0$ for $x$ into $y=5x^2+ax+b$ gives $y=-\frac15$ as our minimum.