If the parabola represented by f(x) = Px^2 +Qx+ R passes thru

The question is testing how well can symmetry be spotted if the x- and y- intercerpts are the only coordinates known. The graph is shifted left, which is proven by the x- intercepts (3,0) and (-5,0) and therefore statements two and three are true.

E.

Thanks,

Hey Guys,

I went thought a litle different path and got different outcomes, would you help me?

Fisrt I tried to solve for P, Q and R so:

25P -5Q+R=0 for points (-5,0)
9P +3Q +R=0 for points (3,0)
and R= -2 for point (0,-2)

So coming back to the first equations:

25P -5Q= 2
9P +3Q=2

This yelds P= 2/15 an Q= -1/15

So the equation would be: 2/15 X^2 -1/15 X -2

If we use F(0) and F(-2) on this equation it woun´t have the same result, where am I wrong?

Thanks A lot

Your value of Q is not correct. If you substitute the above values of P and Q into equation 1 , you will see you do not get 2. As a thumb rule, you should check that the values of the variables that you get by solving systems of equations is actually correct or not.

The correct values are P=2/15 and Q=4/15 (with R=-2). When you substitute these values, you will see that only equations 2 and 3 are correct. Thus, E is the correct answer.

Interesting approach and something that will save you a lot of time. The trick here is to recognize that as the parabola passes through (-5,0) and (3,0), the axis of symmetry for this parabola will be X=-1. This information will come in handy for proving that statement 2 is indeed correct. Statements 1 and 3 are pretty obvious from the graph.

Hello

Can you please elaborate on how can we find how a parabola is symmetrical around a certain axis (if not origin)?

Thanks

For any parabola in standard form equation , y= ax^2+bx+c has equation of axis of symmetry as x=-b/(2a)

Hope this helps.

Plotting the Points on the X-Y Axis for the Parabola:


the 2 X-Intercepts occur at (-5 , 0) and (+3 , 0)

the Y-Intercept occurs at (0 , -2)


This tells us that the Parabola must be an UPWARDS-OPENING Parabola. Thus, the Minimum Value of f(x) will occur at the VERTEX where X = (-b) / 2a

However, since we are NOT Given Values for the Coefficients in the Quadratic Equation of the Parabola, we can use the Concept that the Parabola is SYMMETRIC around the Vertex/Lowest Point of the Upwards Opening Parabola given here.


Since ONE of the X-Intercepts occurs at X = -(5) and the SECOND of the X-Intercepts occurs at X = +3 -

the Vertex MUST occur at the MID-POINT of these TWO X-Intercepts.


(5 Units to the Left + 3 Units to the Right) / 2 = 4 Units


Thus, 4 Units to the RIGHT of (-5 , 0) will give us the Vertex of the Parabola, which will be the MIN Value of f(x).

This Vertex occurs at: When X = -(1) ------ Therefore, the MIN Y Value will occur at F(-1) ----- Any X Value to the RIGHT of -(1) or the LEFT of -(1) will always produce a Greater Y-Value than the Vertex F(-1)

(we could have also found it by moving on the X-Axis 4 Points to the LEFT of (+3 , 0) )


I. F(4) < F(3)

F(4) is FURTHER to the Right of the Vertex Point F(-1) than is F(3)

Therefore, F(4) will give us a Y-Value that appears at a HIGHER POINT on the Parabola.

This means this Statement is FALSE: F(4) > F(3)


II. F(0) = F(-2)

Both X-Values of 0 and -(2) are EXACTLY 1 UNIT away from the MIN Value of Y at the Vertex where X = -(1)

Because the Parabola is SYMMETRIC around the Axis of Symmetry that runs through the Vertex, the Y Value produced by F(0) and F(-2) will give the SAME Y Value

MUST BE TRUE


III. F(3) > F(-1)


MUST BE TRUE --- as was said above, the MIN Value will occur at the Vertex when X =-(1)

Any X Value (such as X = +3) will produce a greater Y-Value than F(-1)

MUST BE TRUE


-E-

2 and 3 must be true

f(

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