Write the equation of a quadratic function given only the vertex

**shadow6** · March 11th 2010, 01:32 PM

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

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

**SuperCalculus** · March 11th 2010, 01:53 PM

Originally Posted by shadow6

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

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

What you'd have to do, is expand that function, to get;

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

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

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

Hence, $0 = 0$

Damn.

**mr fantastic** · March 11th 2010, 02:00 PM

Originally Posted by shadow6

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

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

**shadow6** · March 11th 2010, 02:07 PM

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

**mr fantastic** · March 11th 2010, 02:23 PM

Originally Posted by shadow6

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

No, the parabola has a minimum therefore a > 0, which is not the same as any real number.

**Tim28** · March 12th 2010, 01:47 PM

Originally Posted by shadow6

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

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

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