Question

Four equivalent forms of a quadratic function are given. Which form displays the zeros of function h?

Answer 1

Step-by-step explanation:

As we know that $f$ is a polynomial function, the values of $x$ for which $f(x)=0$ are said to be the zeros of  $f$.

Factoring the polynomial equation, if it is eligible to get factored, setting each factor equal to zero and solving can determine the zeros.

So, it is clear from above that option A i.e.  $h\left(x\right)=-4\left(x\:-\:2\right)\left(x\:+\:2\right)$ and option C i.e. $h\left(x\right)\:=\:4\left(x^2\:-\:4\right)$ displays the zeros of function h, as they can be factored by setting each factor equal to zero.

Option A)

Lets solve them to get the zeros of these functions.

Considering the function

$h\left(x\right)=-4\left(x\:-\:2\right)\left(x\:+\:2\right)$

$\mathrm{The\:roots\:are\:the\:intercepts\:with\:the\:x-axis}\:\left(y=0\right)$

$-4\left(x-2\right)\left(x+2\right)=0$

Using the Zero Factor Principle:

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

So,

$\mathrm{Solve\:}\:x-2=0:\quad x=2$

$\mathrm{Solve\:}\:x+2=0:\quad x=-2$

$\mathrm{The\:zeros\:to\:the\:quadratic\:equation\:are:}$

$x=2,\:x=-2$

Option B)

Considering the function

$h\left(x\right)\:=\:4\left(x^2\:-\:4\right)$

$\mathrm{The\:roots\:are\:the\:intercepts\:with\:the\:x-axis}\:\left(y=0\right)$

$4\left(x^2-4\right)=0$

$\mathrm{Divide\:both\:sides\:by\:}4$

$\frac{4\left(x^2-4\right)}{4}=\frac{0}{4}$

$x^2-4=0$

$x^2=4$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

Although the zeros of the function $h\left(x\right)=-4x^2\:+\:16$ can also be obtained for the function by similar method. But, for this we would have to solve them before determining the zeros.

For example,

Considering the function

$h\left(x\right)=-4x^2\:+\:16$

$\mathrm{The\:roots\:are\:the\:intercepts\:with\:the\:x-axis}\:\left(y=0\right)$

$-4x^2+16=0$

$-4x^2+16-16=0-16$

$-4x^2=-16$

$x^2=4$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

Keywords: zeros, quadratic function

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Answer 2

Answer:

D

Step-by-step explanation: