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anonymous

  • 4 years ago

Find the axis of symmetry for the following function. f(x) = -2x2 - 4x x=?

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  1. anonymous
    • 4 years ago
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    hold on im gonna look in my notes

  2. anonymous
    • 4 years ago
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    well that was wrong. vertex is \[-\frac{b}{2a}=-\frac{4}{4}=-1\]

  3. anonymous
    • 4 years ago
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    how would i find the y-intercept of this problem?

  4. anonymous
    • 4 years ago
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    y = -2(x^2 + 2x) y = -2(x^2 + 2x +1 - 1) *divide the middle number by two then square it (subtract and add that number) y = -2 (x^2 + 2x + 1) +2 *take the subtracted number and multiply it by a to move it out of the brackets y = -2(x + 1) ^2 + 2 *take the middle term and divide it by 2 to get the x value. then square the whole bracket your vertex is (-1, 2) * h value is always opposite of the sign

  5. anonymous
    • 4 years ago
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    x = -1 <----- axis of symmetry as for your y - intercept set all your x values to 0 and solve for y y = -2x^2 - 4x y = -2(0)^2 - 4(0) y = 0 * your y intercept is 0 Note f(x) is y

  6. anonymous
    • 4 years ago
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    y intercept, set it equal to zero and solve get \[-2x^2-4x=0\] \[-2x(x+2)=0\] \[x=0,x=-2\]

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