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Which equation, when graphed, has x-intercepts at (-1,0) and (-5,0) and a y-intercept at (0, -30)?
f(x) = -6(x + 1)(x + 5)
f(x) = -6(x - 1)(x - 5)
f(x) = -5(x + 1)(x + 5)
f(x) = -5(x - 1)(x – 5)

Respuesta :

If a function has roots -1 and -5, it must be in the form

[tex]f(x)=a(x+1)(x+5)[/tex]

We can fix the coefficient a by imposing the passing through (0,-30):

[tex]f(0)=-30=a(1)(5)\iff a=-6[/tex]

So, the function is

[tex]f(x)=-6(x+1)(x+5)[/tex]

abidemiokin

The equation when graphed, has x-intercepts at (-1,0) and (-5,0) and a y-intercept at (0, -30) is f(x) = -4(x+1)(x+5)

Equation of quadratic functions

Quadratic functions has a leading degree of 2. The standard equation is given as:

f(x) = a(x - a)(x - b)

Given the x-intercepts at (-1,0) and (-5,0) will give;

f(x) = a(x+ 1)(x+5)

Given the y-intercept at (0, -30), hence;

-20 = a(1)(5)

-20 = 5a

a = -4

Determine the required equation

f(x) = -4(x+1)(x+5)

Hence the equation when graphed, has x-intercepts at (-1,0) and (-5,0) and a y-intercept at (0, -30) is f(x) = -4(x+1)(x+5)

Learn more on quadratic function here: https://brainly.com/question/1214333

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