J's answer to Tammy Chan's Hong Kong question.
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b)
The graph cuts the y-axis at Q. At any point on the y-axis, x = 0. So at Q, the x-coordinate of the graph is also 0.
Since we now know the equation of the graph to be y = -(x - 4)² + 9, substitute x = 0 to find the y-coordinate at Q.
y = -(0 - 4)² + 9
y = -(4²) + 9
y = -16 + 9 = -7
So the coordinates of Q are (0,-7)
The graph cuts the y-axis at Q. At any point on the y-axis, x = 0. So at Q, the x-coordinate of the graph is also 0.
Since we now know the equation of the graph to be y = -(x - 4)² + 9, substitute x = 0 to find the y-coordinate at Q.
y = -(0 - 4)² + 9
y = -(4²) + 9
y = -16 + 9 = -7
So the coordinates of Q are (0,-7)
Date Posted:
3 years ago
Slight error in the last few steps. It should be
y = -(-4)² + 9 instead of y = -(4²) + 9
y = -(-4)² + 9 instead of y = -(4²) + 9
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