For what values of m does the graph of y = mx2 – 5x – 2 have no x-intercepts?
Accepted Solution
A:
"no x-intercept(s)" => graph never even touches the x-axis.
y = mx^2 – 5x – 2 is the equation of a parabola that opens up.
The quadratic formula would be useful here. Note that there are 3 possible outcomes when this formula is applied: the discriminant b^2 - 4ac could be zero, positive or negative. If the discrim. is negative, then the roots (solutions) are complex; the graph does not touch the x-axis.
So, let's find the discrim. of y = mx2 – 5x – 2. here a=m, b= -5 and c= -2. The discrim. is b^2 - 4(a)(c), or (-5)^2 - 4(m)(-2), or 25+8m. For which values of m is 25+8m<0? Subtracting 25 from both sides, we get 8m < -25, or m<25/8.
Let's check this. Let m=3 (which is less than 25/8). Does the curve ever touch the x-axis? y = 3x^2 - 5x - 2; here a=3, b= -5 and c= -2. Then the discriminant is 25-4(3)(-2) = 49 (which is positive), so there would be two different, real roots (x-intercepts). No good.
Looking at y=mx^2 - 5x - 2 once more, we see that if x=0, y = -2, so for any m, the curve goes thru the point (0,-2), so long as m is positive.
Now let's look at the possibilities for negative m. Suppose we have y = -3x^2 - 5x - 2. Then the discriminant would be 25-4(-3)(-2), or 1. So we'd have 2 real, unequal roots yet again.
As a last resort, I used my calculator repeatedly to graph y=mx^2 - 5x - 2. I found that if m is less than about -3.125, there is/are no x-intercept. (answer)