How would know if the curve y=x^2+2x-7 intercept the line 2x-y-3=0?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to test if the given line `2x - y - 3 = 0` intersects the curve `y = x^2 + 2x - 7` , hence, you need to check if the following system of equations has solutions, such that:

`{(2x - y - 3 = 0),(y = x^2 + 2x - 7):}`

You may use substitution method to evaluate the solution to the system, hence, you may replace `x^2 + 2x - 7` for `y` , in the top equation, such that:

`2x - (x^2 + 2x - 7) - 3 = 0`

`2x - x^2 - 2x + 7 - 3 = 0`

Reducing duplicate members, yields:

`-x^2 + 4 = 0 => x^2 = 4 > x_(1,2) = +-2 => y_(1,2) = (+-2)^2 + 2*(+-2) - 7 => y_1 = 1; y_2 = -7`

Hence, testing if the system of equations `{(2x - y - 3 = 0),(y = x^2 + 2x - 7):}` has solutions yields `(2,1)` and `(-2,-7)` .

Hence, since the system of equations has two solutions, the line `2x - y - 3 = 0` intersects the curve `y = x^2 + 2x - 7` at the points `(2,1)` and `(-2,-7).`

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The curve y=x^2+2x-7 intercepts the line 2x-y-3=0 if the solution of the two equations is a real number.

2x - y - 3 = 0 gives y = 2x - 3

If y = 2x - 3 is substituted in y=x^2+2x-7 the quadratic equation obtained is:

2x - 3 = x^2 + 2x - 7

x^2 - 4 = 0

x^2 = 4

x = +- 2

The roots of the equation are real. As this is the case, the curve y=x^2+2x-7 intersects the line 2x-y-3=0.

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