For (a); I tried this:
y = x^2 + qx + r
y = o^2 + 6 * 0 + 16
y = -16
y = x^2 -6x - 16
y = 8^2 - 6*8 - 16
y = 0
I am lost with (b) and not sure about (c)
Thanks.
For a)
$\displaystyle f(x) = a(x - \alpha)(x - \beta) $
$\displaystyle f(x) = a(x - 8)(x + 2) $
$\displaystyle f(x) = a(x^2 - 6x - 16) $
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For b)
I think we can get the minimum point by using Calculus.
$\displaystyle f(x) = (x^2 - 6x - 16) $
$\displaystyle \frac{dy}{dx} = 2x - 6 = 0 $
$\displaystyle x = 3 $
Set $\displaystyle x = 3 $ into $\displaystyle f(x) = (x^2 - 6x - 16) $
Then $\displaystyle f(3) = -25 $
So point P's coordinates is: (3 ; - 25)
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For c)
Now this one really isn't hard.
Look at this:
Would you be able to solve this?
$\displaystyle k^2 - 6k - 16 = 0 $
Of course you would!
But look at what i did:
I simply treated $\displaystyle (x + 3) $ as $\displaystyle k $
So let's solve this one:
$\displaystyle k^2 - 6k - 16 = 0 $
$\displaystyle (k - 8)(k + 2) = 0 $
$\displaystyle k = 8 \ and \ k = -2 $
But remember that $\displaystyle k $ is actually equal to $\displaystyle (x + 3) $
So $\displaystyle (x + 3) = 8 $ and $\displaystyle (x + 3) = -2 $
$\displaystyle x = 5 $ or $\displaystyle x = -5 $
To (b)
I don't think GAdams has taken a calculus class as yet. We can use the vertex formula to get the lowest Point.
$\displaystyle y = x^2 - 6x - 16$
This is an upward opening parabola, so the vertex occurs at the minimum point.
For vertex, $\displaystyle x = \frac {-b}{2a} = \frac {6}{2} = 3$
when $\displaystyle x = 3$
$\displaystyle y = 3^2 - 6(3) - 16 = -25$
So the coordinates for the lowest point is $\displaystyle (3,-25)$