Does the parabola y=a(x-h)^2+k intersect the x-axis? 1) h

Question

Does the parabola  y=a(x-h)^2+k  intersect the  x -axis?
1)  h<0 
2)  a<0

CrackverbalGMAT · Answer

y = a(x - h)^2 + k is the vertex form of parabola, where h is the x co- ordinate of the vertex and k is the y co-ordinate of the vertex The position of the vertex depends on the values of h and k and the shape of the parabola depends on a If a > 0, then it is an upward parabola and if a < 0, it is a downward parabola Statement (1) h < 0. From this we can gather that the vertex of the parabola can be below or above the x axis. Four possibilities can arise. If h < 0, k < 0 and a < 0, then the vertex is in the 3rd Quadrant and the parabola will be downward and it will not intersect the x axis If h < 0, k > 0 and a < 0, then the vertex is in the 2nd Quadrant and the parabola will be downward and it will intersect the x axis If h < 0, k < 0 and a > 0, then the vertex is in the 3rd Quadrant and the parabola will be upward and it will intersect the x axis If h < 0, k > 0 and a > 0, then the vertex is in the 2nd Quadrant and the parabola will be upward and it will not intersect the x axis Therefore Statement (1) Alone is insufficient as we do not know the values of k and a The answer can be B, C or E Statement (2) a < 0. From this we can gather that the parabola is a downward parabola. Four Possibilities can arise. If a < 0, h > 0 and k > 0, then the vertex is in the 1st Quadrant and the parabola will be downward and it will intersect the x axis If a < 0, h < 0 and k > 0, then the vertex is in the 2nd Quadrant and the parabola will be downward and it will intersect the x axis If a < 0, h < 0 and k < 0, then the vertex is in the 3rd Quadrant and the parabola will be downward and it will not intersect the x axis If a < 0, h > 0 and k < 0, then the vertex is in the 4th Quadrant and the parabola will be downward and it will not intersect the x axis Therefore Statement (2) Alone is insufficient as we do not know the values of h and k The answer can be C or E Combining Statements (1) and (2) h < 0 and a < 0 We understand that this is a downward parabola who's x coordinate is negative. Two possibilities can arise. If a < 0, h < 0 and k > 0, then the vertex is in the 2nd Quadrant and the parabola will be downward and it will intersect the x axis If a < 0, h < 0 and k < 0, then the vertex is in the 3rd Quadrant and the parabola will be downward and it will not intersect the x axis Therefore Statement (3) also is insufficient as we do not have the value of k. Option E Arun Kumar

hiranmay · Answer

Does the parabola y=a(x-h)^2+k intersect the x -axis? 1) h<0 2) a<0 Solution : y=a(x-h)^2+k =>y=0 @intersection point, so 0=a(x-h)^2+k => (x-h)^2 = -k/a to get a real value of x -k/a should be greater than 0 i.e. -k/a > 0 => -ka>0 => ka<0 so k = +ve & a=-ve or k = -ve & a=+ve => we need to know both the type of k & a (1) h<0--> insuff: we don't know anything about k & a i.e. +/- (2) a < 0 --> insuff: we don't know anything about k i.e. +/- combining (1) & (2) also, we don't know anything about k i.e. +/- Answer: E

IanStewart · Answer

For the parabola to meet the x-axis, it would need at least one x-intercept. So if we set y = 0, the equation would need to have a solution. Substituting y = 0, we get:

0 = a(x-h)^2 + k
-k = a(x - h)^2

Using both statements, we know a is negative, and (x-h)^2 is a square, so is positive (or zero). So the right side of the equation above is negative (or zero). But we know nothing about k: if k is negative, then -k is positive. So then -k could not possibly equal the negative (or zero) number a(x-h)^2. So in that case, the parabola does not meet the x-axis. But if k is positive, both sides of the equation can be negative, and we can definitely set values for the unknowns so the equation has solutions.

So the answer is E.

Aulegue · Answer

We can also solve the problem, using the discriminant. I know that it's not a fancy approach but still

A parabola has real root/roots if its discriminant D>=0
f(x)=a(x)^2-2ah*x+a(h)^2+k
D=(a)^2(h)^2-(a)^2(h)^2-ak=-ak

So the question becomes whether ak<=0?

1) tells nothing about a, k - insuff
2) ak<=0? --> k>=0? - insuff

1)&2) we still don't know anything about k

The answer is E.

PS. The vertex of our parabola is the point with coordinates (h,k)
From statements 1&2 we know that the parabola lies somewhere left of the origin and that it's a downward one, but (depending on k) the vertex may be above (two intersections), on (one intersection), or below (no intersections) the x-axis.

Does the parabola y=a(x-h)^2+k intersect the x-axis? 1) h<0 2) a<0

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