The equation be 2x² - 3x - 5 then the value of x = -1.
Let a and b represent the coefficients of a parabola in the standard form ax² + bx +c = 0.
Part A: Let f(x) = 0 and solve for x to get the x-intercepts.
Given the function 2x²-3x - 5 = 0
Factor to get (2x - 5) (x + 1) = 0
then 2x - 5 = 0 and x = 2.5x + 1 = 0 and x = -1
These are the x-intercepts.
Part B: As the coefficient of the [tex]$x^2[/tex] term exists positive, the parabola opens up, and the vertex exists at a minimum. The coordinates of the parabola exist given by -b/2a, f( -b/2a)
where a and b represent the coefficients of a parabola in the standard form ax² + bx +c = 0.
In this case a = 2, b = -3 and c = -5.
So -b/2a = 3 / 4
f(-b/2a) = [tex]2 * (3/4)^2 -3 * (3/4) - 5[/tex] = -6.125.
The coordinates of the vertex exist (3/4, -6.125)
Part C: I will give them a minimum from completing the square
y = [tex]$2(x - 0.75)^2[/tex] - 6.125
as x approaches + ∞, y approaches + ∞.
as x approaches - ∞, y approaches + ∞.
It's a quadratic.
The y values go to + ∞ when x goes from - ∞ to + ∞.
Part D: The graph is as follows:
To learn more about functions refer to:
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