Mathematics, 25.07.2019 02:00 KekePonds1021

For what values of m does the graph of y = mx2 – 5x – 2 have no x-intercepts?

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Answer from: bob7220

The value of m is less than -3.125 does the graph y = mx^2 - 5x - 2 has no x-intercept.

We have to determine, the value of m for does the graph y = mx^2 - 5x - 2 have no intercept.

According to the question,

y = mx^2 - 5x -2 is the equation of a parabola that opens up.

The quadratic formula would be useful here.

There are 3 possible outcomes when this formula is applied: the discriminant b^2 - 4ac could be zero, positive or negative.

If the discriminant is negative, then the roots (solutions) are complex; the graph does not touch the x-axis.

Therefore,

To find the discriminant of graph = y = mx^2 - 5x -2 .

D = b^2-4ac

Where, a = 1, b =-5, c=-2.

Then,

$D = (-5)^2-4\times 1 \times (-2)\\\\D = 25+8\\\\D = 33$

Then, The value of m is 25+8m

Subtracting 25 from both sides,

$8m < -25\\\\m = \dfrac{-25}{8}$

Let, m=3 (which is less than 25/8).

The curve ever touches the x-axis, y = 3x^2 - 5x - 2;

Where 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).

Looking at once more, y=mx^2 - 5x - 2 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, y=mx^2 - 5x - 2 .

Then the discriminant would be = 25-4(-3)(-2) = 1.

They have 2 real, unequal roots yet again.

Hence, The graph y= y=mx^2 - 5x - 2. found that if m is less than about -3.125, there is no x-intercept.

To know more about the Slope click the link given below.

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Answer from: isiahb123

"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)

Answer from: Quest

7.666270253x10 raised to the 18th power

Answer from: Quest

Wow. that is a really cool question.

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