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ser-zykov [4K]
3 years ago
8

Write the equation for a parabola that has x− intercepts (−1.6,0) and (−3.2,0) and y−intercept (0,25.6).

Mathematics
2 answers:
tankabanditka [31]3 years ago
6 0
X-intercepts are zeroes of the parabola.
(x-x1)(x-x2)=(x+1.6)(x+3.2)=x² +1.6x+3.2x+5.12=x²+4.8x+5.12
 
y=x²+4.8x+5.12, but the y-int. is (0, 25.6),

So, our parabola should look like
y=5(x²+4.8x+5.12)
y=5x²+24x+25.6
Bess [88]3 years ago
5 0
 Answer: y = 5(x + 1.6)(x + 3.2)

∞∞∞∞∞∞∞∞
Explanation:
∞∞∞∞∞∞∞∞

<em>x-intercept = (-1.6, 0) and (-3.2, 0)</em>

=====>   y = a(x + 1.6)(x + 3.2)

<em>Given y-intercept = (0, 25.6)</em>

=====>  25.6 = a( 0 + 1.6)(0 + 3.2) 
=====>  25.6 = 5.12a
=====>  a = 5

<em>Plug in a = 5 into the equation
</em>
=====>  y = 5(x + 1.6)(x + 3.2)
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Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

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