Can someone help me with a question!?!?

Mathematics

1 answer:

Len [333]2 years ago

8 0

If the parent graph f(x) = x² is changed to f(x) = 2x², the vertex of the parabola will still remain (0, 0) because whenever the equation of a parabola is in the form y = ax², the vertex will always be (0, 0).

Now if <em>a</em> is a big number, the parabola will become narrower.

So it will stretch vertically and become narrower.

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Find a polynomial function of least degree having only real coefficients, a leading coefficient of 1, and roots of 2-√6 , 2+ √6

horsena [70]

Answer:

$x^{4} -18x^{3}+104x^{2} -172x-100$

Step-by-step explanation:

The 3 roots are given out of which 2 are real and 1 is imaginary. For a polynomial of least degree having real coefficients, it must have a complex conjugate root as the 4th root. Therefore, based on 4 roots, the least degree of polynomial will be 4. Finding the polynomial having leading coefficient=1 and solving it based on multiplication of 2 quadratic polynomials, we get:

$\\\\x_{1} = 2-\sqrt{6} \\x_{2} = 2+\sqrt{6} \\x_{3}=7-i \\x_{4}=7+i \\\\P(x)=1(x-x_{1})(x-x_{2} )(x-x_{3} )(x-x_{4} ) \\\\=(x-(2-\sqrt{6}))( x-(2+\sqrt{6} )) (x-(7-i))( x-(7+i))\\=((x-2)+\sqrt{6})( ( x-2)-\sqrt{6} ) ((x-7)+i)( (x-7)-i)\\=((x-2)^{2} -(\sqrt{6} )^{2} )((x-7)^{2}-(i)^{2})\\=(x^{2} -4x-2)(x^{2} -14x+50)\\=x^{4} -18x^{3}+104x^{2} -172x-100\\$