4 Answers | Add Yours
y = 2x^2 - 3x -20
we need to determine x-intercept.
x-intercept is the point where the function intersect with the x-axis. In this case y value will be zero.
Then we need to calculate x values for the functin y to be zero.
y = 2x^2 - 3x - 20 = 0
==> (2x +5)(x-4) = 0
==> x1= -2/5
==> x2= 4
Then y intersect with x-axis in two point:
(-2/5, 0) and (4, 0)
The x intercepts are, in fact, the roots of the equation:
2x^2 - 3x – 20 = 0 (when the graph of the y function is intercepting x axis).
Now, all we have to do is to solve the equation.
Because it's a quadratic, we'll apply the quadratic formula:
x1 = [-b+sqrt(b^2 - 4ac)]/2a
x2 = [-b-sqrt(b^2 - 4ac)]/2a
We'll identify a, b and c, which are the coefficients of the function:
y = ax^2 + bx + c
y = 2x^2 - 3x – 20
a = 2, b = -3, c = -20
Now, we'll calculate the roots:
x1 = [-(-3) + sqrt(9 +160)]/2
x1 = (3+13)/2
x1 = 8
x2 = (3-13)/2
x2 = -5
So, the x intercepts are x1 and x2.
y = 2x^2-3x-20.
To find the x intercepts, find values of x for which y = 0.
So 2x^2-3x-20 = 0.
We factorise 2x^2-3x-20 by splitting the middle term , -3x into -8x + 5x and grouping:
2x^2 - 8x + 5x - 20 = 0
2x(x-4) +5(x-4) = 0
(x-4)(2x+5) = 0
x-4 = 0 or 2x+5 = 0.
x-4 = 0 gives x = 4.
2x+5 = 0 gives x = -5/2
So the x intercepts of y = 2x^2-3x-20 = 0 are at x = 4 and x= -5/2 on the x axis.
To find the x-intercepts of y = 2x^2 - 3x - 20, we have to equate y = 0 and solve the resulting quadratic equation.
So if we substitute y=0 for the equation y = 2x^2 - 3x - 20 we get :
2x^2 - 3x - 20 = 0
=> 2x^2 - 8x + 5x - 20 = 0
=> 2x (x-4) + 5 (x-4) = 0
=> (2x+5) (x-4) = 0
=> x = -5/2 or x = 4
Therefore the x - intercepts are at x = -5/2 and at x = 4 .
We’ve answered 319,852 questions. We can answer yours, too.Ask a question