Given axis of symmetry$\ x=1$ for the parabola $\ y = ax^2 + 4x - 2$. What is the value of $ a$?
I am just curious what is proper method of solving this. This is how I did it: $$ y = ax^2 + 4x - 2 $$ $$ 0 = a(1)^2 + 4(1) - 2 $$$$0 = a + 2 $$$$a = -2$$
Is this the correct way of solving it? Are there any other methods?
On
The axis of symmetry of a parabola, is essentially the unique point $p$ such that $f(p+a) = f(p-a)$ for all $a \geq 0$, since this is what symmetry means about that point. Here we are given $p = 1$.
Now all we need to do, is substitute this in : $$f(1+x) = f(1-x) \implies a(1+x)^2 + 4(1+x) - 2 = a(1-x)^2 + 4(1-x) - 2$$
Now, simplify the above to get: $$ a(1+x)^2 =a(1-x)^2 - 8x \implies 4ax = -8x $$
Since this is true for all $x \geq 0$, you can conclude that $4a = -8$ and hence $a = -2$.
On
There exists a form in which to write quadratic equations called the turning point form, or completing the square. $$ y=a(x-h)^2+k $$ where $a$ is the horizontal dilation, $(h,k$) is the turning point. This implies the axis of symmetry is at $x = h$.
For your above example, you would solve the question like this:
\begin{align} y & = ax^2 + 4x - 2 \\ & = a(x^2+\frac{4}{a}x - \frac {2}{a}) \\ & = a[(x+\frac{2}{a})^2-\frac{4}{a^2}-\frac{2}{a}] \\ & = a[(x+\frac{2}{a})^2-\frac{4-2a}{a^2}] \\ & = a(x+\frac{2}{a})^2-\frac{4-2a}{a} \\ \end{align}
We know axis of symmetry is at $x=1$, and we have previously discussed that $h$ is the axis of symmetry of the turning point form. Therefore it is correct to say $h=-\frac{2}{a}$.
\begin{align} h &= -\frac {2}{a}\\ \therefore a &= -2 \end{align}
No, you are claiming that the quadratic curve passes through $(1,0)$ without justification.
Guide:
Notice that the extreme point occur at axis of symmetry for a quadratic function. Hence if you differentiate the quadratic function and equate it to zero, $x=1$ is a solution.
Alternatively, you can also obtain the axis of symmetry by using completing the square technique.