How do you write the quadratic function in standard form $y = - \frac{1}{2} (x - 6) (x - 8)$ $y=-1/2(x-6)(x-8)$ ?

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How do you write the quadratic function in standard form $y = - \frac{1}{2} (x - 6) (x - 8)$ $y=-1/2(x-6)(x-8)$ ?

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Answer 1 · 2017-04-03T14:14:08

$7 x - \frac{x^{2}}{2} - 24$ $7x-x^2/2-24$

Explanation:

Let's first multiply $(x - 6) (x - 8)$ $(x-6)(x-8)$ . We will use FOIL to expand this (first, outer, inner, last).

First: $x \cdot x = x^{2}$ $x*x=x^2$
Outer: $x \cdot - 8 = - 8 x$ $x*-8=-8x$
Inner: $x \cdot - 6 = - 6 x$ $x*-6=-6x$
Last: $- 8 \cdot - 6 = 48$ $-8*-6=48$

Add these all up to get $x^{2} - 14 x + 48$ $x^2-14x+48$ .

Now, don't forget to multiply each of these terms by $- \frac{1}{2}$ $-1/2$ to get $- \frac{x^{2}}{2} + 7 x - 24$ $-x^2/2+7x-24$ . We can arrange the terms so that the positive term is in the front $7 x - \frac{x^{2}}{2} - 24$ $7x-x^2/2-24$ .

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How do you write the quadratic function in standard form $y = - \frac{1}{2} (x - 6) (x - 8)$ $y=-1/2(x-6)(x-8)$ ?

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How do you write the quadratic function in standard form $y = - \frac{1}{2} (x - 6) (x - 8)$ $y=-1/2(x-6)(x-8)$ ?