Determine if the quadratic functions are positive definite, negative definite or neither:
a. y=-2x^2+3x+2
b. 3x^2+x+11
Please show work, thank you!
a. opens down; discriminant is positive, so neither (it has real roots)
b. opens up; discriminant is negative, so it does not cross the x-axis -- positive definite
read up on the discriminant and what it means
To determine whether a quadratic function is positive definite, negative definite, or neither, we need to examine the leading coefficient (the coefficient of the x² term).
a. y = -2x² + 3x + 2
The leading coefficient is -2. Since it is negative, we know that the quadratic function opens downwards. To determine positive definiteness or negative definiteness, we need to evaluate the discriminant of the function.
The discriminant, denoted by Δ, can be calculated using the formula Δ = b² - 4ac, where a, b, and c are the coefficients of the quadratic equation (in the form ax² + bx + c).
For the function y = -2x² + 3x + 2, the coefficients are:
a = -2
b = 3
c = 2
Substituting these values into the discriminant formula, we have:
Δ = (3)² - 4(-2)(2)
Δ = 9 + 16
Δ = 25
If Δ > 0, the quadratic function is neither positive definite nor negative definite. If Δ < 0, the quadratic function is either positive definite or negative definite. If Δ = 0, the quadratic function is neither positive definite nor negative definite. In this case, since Δ = 25 (which is greater than 0), the quadratic function is neither positive definite nor negative definite.
b. y = 3x² + x + 11
The leading coefficient is 3. Since it is positive, we know that the quadratic function opens upwards. Again, let's calculate the discriminant.
For the function y = 3x² + x + 11, the coefficients are:
a = 3
b = 1
c = 11
Substituting these values into the discriminant formula, we have:
Δ = (1)² - 4(3)(11)
Δ = 1 - 132
Δ = -131
Since Δ = -131 (which is less than 0), the quadratic function is negative definite.