SOLUTION: Give the axis of symmetry and the x-intercepts: Y = 2x^2 - 8x - 42
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Question 49925
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Give the axis of symmetry and the x-intercepts:
Y = 2x^2 - 8x - 42
Answer by
rapaljer(4671)
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The line of symnmetry will be at the vertex. The vertex for y = ax^2 + bx +c will always be at
In this case a=2 and b = -8, so vertex is at
To find the x-intercepts, let y = 0, and
solve the equation 0=2x^2 - 8x - 42 by factoring if possible.
2(x^2 - 4x -21)=0
2(x-7)(x+3)=0
x= 7 ; x = -3
This is a parabola that opens upward. You didn't ask to graph it, but it should look like this:
R^2 at SCC