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How do I do 1 a&b ?

How do I do 1 a&b ?-example-1

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Answer:


Explanation:

Hello!

1a) x(x-4) = 0. To find the y-intercepts, you set x=0 and x-4=0 and then solve for the two x-intercepts: (0,0) and (4,0). The axis of symmetry is halfway between these points: x = 2. Find the value of x(x-4) at x = 2: y = 2(2-4) = -4. Thus, the vertex is (2, -4). Plot these 3 points (0,0), (4,0) and (2, -4) and draw a smooth curve through them. The result is a parabola that opens up.

1b) x^2 + 8x + 12 has the graph of a parabola that opens up. Here the coefficients of the x terms are a = 1, b = 8 and c = 12. The axis of symmetry is x = -b/(2a), which here is x = -8/[2(1)], or x= -4. Substituting -4 for x in x^2 + 8x + 12 results in 16 - 32 + 12, or -4. Thus, the vertex is at (-4,-4).

Next, let y = 0 and find the x-values that satisfy this: x^2 + 8x + 12 = 0.

Let's apply the quadratic formula to solve this for the zeros / roots. We have already identified the coefficients a, b and c (see above). Thus,

-8 plus or minus √(8^2-4(1)(12) ) -8 plus or minus √16

x = ---------------------------------------------- = ----------------------------------

2(1) 2

and so the roots/zeros/solutions are x = -2 and x = -12/2 = -6. Note that the axis of symmetry, x = -4, is exactly halfway between these roots.

We have essentially solved this problem; the roots are -2 and -6. Without more info it's hard to do this problem using the graph of the function, because we don't have other points thru which to draw the graph.



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