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3 years ago

How do I factor the quadratic equation

Mathematics

1 answer:

anastassius [24]3 years ago

4 0

Factor the following:
10 y^2 - 35 y + 30
Factor 5 out of 10 y^2 - 35 y + 30:
5 (2 y^2 - 7 y + 6)
Factor the quadratic 2 y^2 - 7 y + 6.
The coefficient of y^2 is 2 and the constant term is 6.
The product of 2 and 6 is 12.
The factors of 12 which sum to -7 are -3 and -4. So 2 y^2 - 7 y + 6 = 2 y^2 - 4 y - 3 y + 6 = y (2 y - 3) - 2 (2 y - 3):
5 y (2 y - 3) - 2 (2 y - 3)
Factor 2 y - 3 from y (2 y - 3) - 2 (2 y - 3):
Answer: 5 (2 y - 3) (y - 2)

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3 years ago

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mina [271]

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3 years ago

If the parent function f(x) = x^2 is fatter translated 11 units to the left, then translated 5 units down, write the resulting f

nikklg [1K]

The quadratic function given by:

$f(x)=a(x-h)^2+k, \ \ \ a\neq 0$

is in vertex form. The graph of $f$ is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $(h, k)$ . So:

To translate the graph of a function to the right, left, upward or downward we have:

$For \ a \ positive \ real \ number \ c. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:\\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ g(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ g(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{right}: \\ g(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{left}: \\ g(x)=f(x+c)$

By knowing this things, we can solve our problem as follows:

FIRST.

Translating 11 units to the left:

$g(x)=f(x+11) \\ \\ \therefore g(x)=(x+11)^2$

Then translating 5 units down:

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x+11)^2-5$

Since the new function is fatter, the factor we need to multiply the term $(x+11)^2$ must be less than 1, to make the graph fatter. So, according to our options, there are two factors 1/2 and 2.