Does the curve y=b(x-2)^2+c lie completely above the x-axis?

Good question.

A graph will not take a value of x only if it gives an imaginary value of y for that value of x. Otherwise, it will take all values of x.
A line y = mx + c will have some defined value of y for every value of x.
The same isn't true for say a circle or a parabola.
x^2 + y^2 = 1 is a circle of radius 1 with centre (0, 0). This means the graph will exist for x values only from x = -1 to x = 1. When x = 2, y = sqrt(-3) which is imaginary and hence not defined. The graph does not exist for this value.

But in our original question, when x = 2, y = c, a real value. So we know that the graph does exist, does have a y value for x = 2. Whether you draw the whole graph or show only a small section, doesn't matter. It does exist for x = 2.

aditijain1507

For me the most important thing about parabolas is to visualize a rough graph of the parabola in question. If you know how to do that, this is no more than a 30 sec problem. You don't need to calculate the discriminant of the equation.
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CONCEPT

Consider an equation \(y = a(x+b)^2 + c\)
It is important to understand the role each constant plays in the shape of the graph. Once you understand this, any such question will become very easy.

'a' tells you two things
1. Whether the curve will be upward facing or downward facing (If a is negative, then the curve will be downward facing and if it is positive, then the curve will be upward facing)
2. How steep the curve would be (greater the value, steeper the curve)

'b' tells you the position of the curve relative to the x-axis i.e. whether it will lie to the right of the y-axis or to the left of the y-axis or will be on the y-axis (symmetric relative to the y-axis). Just to be clear, by saying "where the curve lies", I mean where the lowest point of the graph is.
1. A negative value of b means that the lowest point of the curve would lie in the half of the graph where x is positive i.e. either in Quadrant 1 or in Quadrant 4.
2. A positive value of b means that the lowest point of the curve would lie in the half of the graph where x is negative i.e. either in Quadrant 2 or in Quadrant 3.

'c' tells you the position of the curve relative of the y-axis i.e. whether it will lie above the x-axis or on the x-axis or below the x-axis.
1. A negative value of c means that the lowest point of the curve would lie in the half of the graph where y is negative or the lowest point on the curve is below the y-axis.
2. A positive value of c means that the lowest point of the curve would lie in the half of the graph where y is positive or the lowest point on the curve is above the y-axis.
3. If c is 0 then the curve will lie on the x-axis.

With that understood, let's get back to the question.
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Does the curve y=b(x−2)2+cy=b(x−2)2+c lie completely above the x-axis?

Statement 1: b>0,c<0
Statement 2: b>2,c<2

The question asks whether the curve touches the x-axis.
Now, if you have followed the aforementioned concept, then (x-2) doesn't have a role to play in the question.

Statement 1: b>0 and c<0
The graph would roughly look like in attachment 1.
The graph will be upward facing and would lie below the x-axis and would touch the x-axis at two points.
Sufficient

Statement 2: b>2 and c<2
C could be positive or it could be negative
In one case it would touch the x-axis and in the other it won't. So, insufficient.

So, A

chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmat

Dear experts: Does GMAT test the concepts of conic sections?

Regards,
Arup

Hi Vertiaskarishma
I am not very clear.

y = (positive or zero) + c

Since there is no limitation/restriction of c given. If absolute of c is a small number, y can be positive.
If absolute of c is a large number, y can be negative.

Pls let me know where am I going wrong.

The curve is of the standard-form: \(Ax^2 + Bx + C\)
Discriminant (D): \(B^2 - 4AC\)

The equation holds the lowest/highest point at coordinates \((x,y)\): \(( -B/2A , -D/4A)\)

We need to confirm whether the quadratic equation opens down or up - Refer the picture.

Given:

"completely above x axis" means y is not 0 and not negative. y must be positive.
\(Y = b(x-2)^2+c\)

\(Y = bx^2 - 4bx + 4b + c\)
Comparing with the standard-form:

A: \(b\)
B: \(-4b\)
C: \(4b+c\)

Thus, the Y coordinate having the lowest/highest point: \((-D/4A)\)
Resultant-Discriminant (D): \(- [(-4b)^2 - 4*b*(4b+c) ]/4b\) = \(c\) - Rest of the term cancels out.

Hence, only the value of \(c\) is vital to determine the coordinates.

If \(b > 0\): Parabola opens UP - b is the co-efficient of \(x^2\).

\(c > 0\) ---------> The lowest Y-coordinate is +ve
\(c < 0\) ---------> The lowest Y-coordinate is -ve

If \(b < 0\): Parabola opens DOWN

\(c > 0\) ---------> The highest Y-coordinate is +ve
\(c < 0\) ---------> The highest Y-coordinate is -ve

Thus, Only statement-1 is SUFFICIENT.
Statement-2 can yield +ve, -ve OR zero, depending on the respective values of \(c\)

Could you please explain how you got the discriminant?

Regards,

Ritvik

The curve is of the standard-form: \(Ax^2+Bx+C\)
Discriminant (D): \(B^2−4AC\)
Discriminant (D) = Square of Co-effecient-of \(x\) - 4*( Co-effecient-of \(x^2\) )*( Constant term )

A: Co-effecient-of \(x^2\)
B: Square of Co-effecient-of \(x\)
C: Constant term

Analogically, in the equation: y = b\(x^2\) − 4b\(x\) + 4b+c

A: Co-effecient-of \(x^2\) ---------> \(b\)
B: Square of Co-effecient-of \(x\) --------> \((-4b)^2\)
C: Constant term ----------> \(4b+c\)

Thus, the discrimimant = \(16b^2 - 4b(4b + c) = -4bc\)

Experts have commented on the traditional approach i will go with a more logical less therotical approach
The key to evaluvating the same is checking out whether y tends to zero
and if y what other parametrs holds it that it doesn't tend to zero
modifying the equation
=>2+(y-c/b)^1/2

statement 1:
b>0,c<0
then the square root will have positive value as C cannot cancel out Y since C takes a negative value
b cannot effect since fractions cannot tend to 0 it becomes non defined there
Hence sufficient

statement 2 :
b>2,c<2
This may or may not lead to zero since y can take value 2
hence Insuff
therefore IMO A

But what if x is never equal to 02? how can we be sure that x=2? KarishmaB

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