At how many points does line L intersect with the parabola represented by y = x² - 3x + 6?Generally a parabola and a line can have 0, 1, or 2 intersection points.
Notice than in our case, since the coefficient of x^2 is positive, then we have an upward parabola, which intersects y-axis at (0, 6). Just to illustrate, below is the graph of y = x^2 - 3x + 6:
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(1) Line L is parallel to x-axis. All three cases are possible: line L can be below the given parabola (no intersection), line L can go through the vertex (1 intersection) or cross it above the vertex (2 intersections). Not sufficient.
Notice that if it were: line L is parallel to y-axis then we would know that it would cross a parabola at one point, hence in this case the statement would be sufficient.
(2) Line L passes through the point (0, 16). Since (0, 16) is higher than y-intercept of parabola, then line L is "trapped" and must have at least one intersection with parabola (it'll have only one for example if L is x=0). Not sufficient.
(1)+(2) Line L intersect with the parabola at two points. Sufficient.
Answer: C.
Hi Bunuel ,Can you explain preferably thru diagram how line intersects parabola in 0 and 1 way. I am ok with 2 but not 0 and 1. Please explain.