Question

What are the x-intercepts of the function f(x) = –2x2 – 3x + 20? (–4, 0) and five-halves five-halves and (4, 0) (–5, 0) and (2, 0) (–2, 0) and (5, 0)

Answer 1

Answer:

$\text{x-intercepts are }(-4,0)\text{ and }(\frac{5}{2},0)$

Step-by-step explanation:

$\text{Let y}=-2x^2-3x+20}$

$\text{put y=0}$

$\implies\:-2x^2-3x+20=0$

$\implies\:2x^2+3x-20=0$

$\implies\:2x^2+8x-5x--20=0$

$\implies\:2x(x+4)-5(x+4)=0$

$\implies\:(2x-5)(x+4)=0$

$\implies\:x=-4,\frac{5}{2}$

$\therefore\:\text{x-intercepts are }(-4,0)\text{ and }(\frac{5}{2},0)$

Answer 2

The x-intercepts of the function f(x) are (–4, 0) and (5/2,0).

Step-by-step explanation:

The given function is

$f(x)=-2x^2-3x+20$

We need to find the x-intercepts of the function.

Splitting the middle term we get

$f(x)=-2x^2-8x+5x+20$

$f(x)=-2x(x+4)+5(x+4)$

$f(x)=(x+4)(-2x+5)$

Equate f(x)=0, to find the x-intercepts.

$(x+4)(-2x+5)=0$

Using zero product property we get

$x+4=0\Rightarrow x=-4$

$-2x+5=0\Rightarrow x=\frac{5}{2}$

Therefore, the x-intercepts of the function f(x) are (–4, 0) and (5/2,0).

#Learn more

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