Find the equation of the line passing through the point (-6,4) that is perpendicular to the line 5x 3y
2 answers:
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Remark
It's asking you to complete the square to find the vertex.
Solve
y = -2x^2 + x + 3 Put brackets around the first 2 terms.
y = (-2x^2 + x) + 3 Pull out the 2.
y = -2(x^2 - 1/2 x) + 3 Be especially observant of the sign on 1/2x Divide the middle term's number by 2 and square it.
y = -2(x^2 - 1/2 x + [(1/2) / 2]^2 ) + 3 [1/2 divided by 2 is 1/4. (1/4)^2 = 1/16; 2*(1/16) = 1/8
y = - 2(x^2 - 1/2 x + (1/4)^2 ) + 3 + 1/8
y = -2(x - 1/4)^2 + 3 1/8
Comment
Answer to your question. The maximum value (3 1/8) is above the y axis. That means the graph of the equation crosses the x axis twice. I'm going to include the graph because in answering the question, there is no harm in seeing the graph.
It's asking you to complete the square to find the vertex.
Solve
y = -2x^2 + x + 3 Put brackets around the first 2 terms.
y = (-2x^2 + x) + 3 Pull out the 2.
y = -2(x^2 - 1/2 x) + 3 Be especially observant of the sign on 1/2x Divide the middle term's number by 2 and square it.
y = -2(x^2 - 1/2 x + [(1/2) / 2]^2 ) + 3 [1/2 divided by 2 is 1/4. (1/4)^2 = 1/16; 2*(1/16) = 1/8
y = - 2(x^2 - 1/2 x + (1/4)^2 ) + 3 + 1/8
y = -2(x - 1/4)^2 + 3 1/8
Comment
Answer to your question. The maximum value (3 1/8) is above the y axis. That means the graph of the equation crosses the x axis twice. I'm going to include the graph because in answering the question, there is no harm in seeing the graph.