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slavikrds [6]
2 years ago
15

What are the exact steps to solve for the x-intercepts of f(x)= −4x2 + 4x + 8? What about −9 = x2 + 6x? Please explain thoroughl

y
Mathematics
1 answer:
victus00 [196]2 years ago
3 0

Answer:

x-intercepts:  (-1, 0) and (2, 0)

x-intercept:  (-3, 0)

Step-by-step explanation:

Given <u>quadratic function</u>:

f(x)=-4x^2+4x+8

The <u>x-intercepts</u> of a <u>quadratic function</u> are the points at which the curve <u>crosses the x-axis</u> ⇒ when y = 0

Therefore, to find the x-intercepts of the given function, set the function to zero:

\implies -4x^2+4x+8=0

Factor out -4:

\implies -4(x^2-x-2)=0

Divide both sides by -4

\implies x^2-x-2=0

Rewrite the middle term as -2x + x:

\implies x^2-2x+x-2=0

Factor the first two terms and the last two terms separately:

\implies x(x-2)+1(x-2)=0

Factor out the common term (x - 2):

\implies (x+1)(x-2)=0

<u>Zero Product Property</u>:  If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the <u>Zero Product Property</u>, set each factor equal to zero and solve for x (if possible):

\begin{aligned}(x+1) & = 0 & \quad \textsf{ or } \quad \quad (x-2) & = 0\\\implies x & = -1 & \implies x & = 2\end{aligned}

Therefore, the x-intercepts are (-1, 0) and (2, 0).

---------------------------------------------------------------------------------------

Given <u>quadratic equation</u>:

-9 = x^2 + 6x

Add 9 to both sides:

\implies x^2+6x+9=0

Rewrite the middle term as 3x + 3x:

\implies x^2+3x+3x+9

Factor the first two terms and the last two terms separately:

\implies x(x+3)+3(x+3)=0

Factor out the common term (x + 3):

\implies (x+3)(x+3)=0

\implies (x+3)^2=0

Square root both sides:

\implies (x+3)=0

Solve for x:

\implies x=-3

Therefore, the x-intercept is (-3, 0).

As the function has a <u>repeated factor</u> (multiplicity of two), the curve will touch the x-axis at (-3, 0) and bounce off.  

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(see attached for reference)

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