1. Equation Graph

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.

2. For a)

$f(x) = a(x - \alpha)(x - \beta)$
$f(x) = a(x - 8)(x + 2)$
$f(x) = a(x^2 - 6x - 16)$

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For b)

I think we can get the minimum point by using Calculus.

$f(x) = (x^2 - 6x - 16)$

$\frac{dy}{dx} = 2x - 6 = 0$

$x = 3$

Set $x = 3$ into $f(x) = (x^2 - 6x - 16)$

Then $f(3) = -25$

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?

$k^2 - 6k - 16 = 0$

Of course you would!

But look at what i did:

I simply treated $(x + 3)$ as $k$

So let's solve this one:

$k^2 - 6k - 16 = 0$

$(k - 8)(k + 2) = 0$

$k = 8 \ and \ k = -2$

But remember that $k$ is actually equal to $(x + 3)$

So $(x + 3) = 8$ and $(x + 3) = -2$

$x = 5$ or $x = -5$

3. 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.

$y = x^2 - 6x - 16$

This is an upward opening parabola, so the vertex occurs at the minimum point.

For vertex, $x = \frac {-b}{2a} = \frac {6}{2} = 3$

when $x = 3$

$y = 3^2 - 6(3) - 16 = -25$

So the coordinates for the lowest point is $(3,-25)$