How to find the x-intercept of a parabola?

Question

The instructions say to find the x and y intercepts of a parabola, and if necessary, round to the nearest tenth. The problem is: 

y= 2x^2 +5x - 42 

I know to find the x intercept you put 0 in the place of y, and so far I have this: 

2x^2 +5x -42 = 0. But where do I go from here? Should I use the quadratic formula?

Answer 1

The function is y = 2x ^2 + 5x - 42.

To find the x - intercept we put 0 in the place of x.

2x ^2 + 5x - 42 = 0

2x ^2 + 5x - 42 is a quadratic, so use quadratic formula to find the roots of the related quadratic equation.

x = [ - b ± √ (b ^2 - 4ac ) ] / 2a.

Compare the equation with standard form of the quadratic equation ax ^2 + bx + c = 0.

a = 2, b = 5, and c = - 42.

Substitute the values of a = 2, b = 5, and c = - 42.

x = [ - 5 ± √ (5^2 - 4(2)(- 42)) ] / 2(2)

   =  [ - 5 ± √ (25 + 336) ] / 4

   = [ - 5 ± √ 361 ] / 4

  = [ - 5 ± 19 ] / 4

  = (- 5 + 19) / 4 and (- 5 - 19) / 4

  = 14 / 4 and - 24 / 4

x = 7 / 2 and - 6.

Therefore the x - intercepts of the given parabola are - 6 and 7 / 2.