# Thread: Write the equation of a quadratic function given only the vertex

1. ## Write the equation of a quadratic function given only the vertex

The function has a minimum value of 6 at x = 4.

I know that part of the equation will be $\displaystyle y = a(x - 4) + 6$, though I have no idea how to get $\displaystyle a$. The answer is "$\displaystyle y = 3(x - 4)^2 +6$; answers may vary". The only thing I can think of would be to create a table of values to input an $\displaystyle x$ and $\displaystyle y$ value. Though I am unsure of how to do that. Any suggestions?

The function has a minimum value of 6 at x = 4.

I know that part of the equation will be $\displaystyle y = a(x - 4) + 6$, though I have no idea how to get $\displaystyle a$. The answer is "$\displaystyle y = 3(x - 4)^2 +6$; answers may vary". The only thing I can think of would be to create a table of values to input an $\displaystyle x$ and $\displaystyle y$ value. Though I am unsure of how to do that. Any suggestions?
What you'd have to do, is expand that function, to get;

$\displaystyle y = (ax - 4a)(x - 4) + 6$

$\displaystyle y = ax^2 - 8ax + 16a + 6$

Then equate to $\displaystyle y = 6$ and $\displaystyle x = 4$.
$\displaystyle 6 = 16a - 32a + 16a + 6$

Hence, $\displaystyle 0 = 0$

Damn.

The function has a minimum value of 6 at x = 4.

I know that part of the equation will be $\displaystyle y = a(x - 4){\color{red}^2} + 6$, though I have no idea how to get $\displaystyle a$. The answer is "$\displaystyle y = 3(x - 4)^2 +6$; answers may vary". The only thing I can think of would be to create a table of values to input an $\displaystyle x$ and $\displaystyle y$ value. Though I am unsure of how to do that. Any suggestions?
Note the correction in red.

Without further information the value of a cannot be got. The only thing that can be said is that a > 0. So either you have not posted all of the question or the question is incomplete.

4. Just read the answers to the succeeding question. Basically, a can be any real number; hence why answers may vary.

I know that part of the equation will be $\displaystyle y = a(x - 4) + 6$, though I have no idea how to get $\displaystyle a$. The answer is "$\displaystyle y = 3(x - 4)^2 +6$; answers may vary". The only thing I can think of would be to create a table of values to input an $\displaystyle x$ and $\displaystyle y$ value. Though I am unsure of how to do that. Any suggestions?