Question

Given f(x)=2x^2+3x-5 for what values of x is f(x) positive

Answer 1

Answer:

The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)

Step-by-step explanation:

we have

$f(x)=2x^{2}+3x-5$

This is a vertical parabola open upward (the leading coefficient is positive)

The vertex is a minimum

The coordinates of the vertex is the point (h,k)

step 1

Find the vertex of the quadratic function

Factor the leading coefficient 2

$f(x)=2(x^{2}+\frac{3}{2}x)-5$

Complete the square

$f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-5-\frac{9}{8}$

$f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-\frac{49}{8}$

Rewrite as perfect squares

$f(x)=2(x+\frac{3}{4})^{2}-\frac{49}{8}$

The vertex is the point (-\frac{3}{4},-\frac{49}{8})

step 2

Find the x-intercepts (values of x when the value of f(x) is equal to zero)

For f(x)=0

$2(x+\frac{3}{4})^{2}-\frac{49}{8}=0$

$2(x+\frac{3}{4})^{2}=\frac{49}{8}$

$(x+\frac{3}{4})^{2}=\frac{49}{16}$

take the square root both sides

$x+\frac{3}{4}=\pm\frac{7}{4}$

$x=-\frac{3}{4}\pm\frac{7}{4}$

$x_1=-\frac{3}{4}+\frac{7}{4}=1$

$x_2=-\frac{3}{4}-\frac{7}{4}=-2.5$

therefore

The function f(x) is negative in the interval (-2.5,1)

The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)

see the attached figure to better understand the problem