Answer:
x-intercepts: (-1, 0) and (2, 0)
x-intercept: (-3, 0)
Step-by-step explanation:
Given quadratic function:
[tex]f(x)=-4x^2+4x+8[/tex]
The x-intercepts of a quadratic function are the points at which the curve crosses the x-axis ⇒ when y = 0
Therefore, to find the x-intercepts of the given function, set the function to zero:
[tex]\implies -4x^2+4x+8=0[/tex]
Factor out -4:
[tex]\implies -4(x^2-x-2)=0[/tex]
Divide both sides by -4
[tex]\implies x^2-x-2=0[/tex]
Rewrite the middle term as -2x + x:
[tex]\implies x^2-2x+x-2=0[/tex]
Factor the first two terms and the last two terms separately:
[tex]\implies x(x-2)+1(x-2)=0[/tex]
Factor out the common term (x - 2):
[tex]\implies (x+1)(x-2)=0[/tex]
Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).
Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):
[tex]\begin{aligned}(x+1) & = 0 & \quad \textsf{ or } \quad \quad (x-2) & = 0\\\implies x & = -1 & \implies x & = 2\end{aligned}[/tex]
Therefore, the x-intercepts are (-1, 0) and (2, 0).
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Given quadratic equation:
[tex]-9 = x^2 + 6x[/tex]
Add 9 to both sides:
[tex]\implies x^2+6x+9=0[/tex]
Rewrite the middle term as 3x + 3x:
[tex]\implies x^2+3x+3x+9[/tex]
Factor the first two terms and the last two terms separately:
[tex]\implies x(x+3)+3(x+3)=0[/tex]
Factor out the common term (x + 3):
[tex]\implies (x+3)(x+3)=0[/tex]
[tex]\implies (x+3)^2=0[/tex]
Square root both sides:
[tex]\implies (x+3)=0[/tex]
Solve for x:
[tex]\implies x=-3[/tex]
Therefore, the x-intercept is (-3, 0).
As the function has a repeated factor (multiplicity of two), the curve will touch the x-axis at (-3, 0) and bounce off.
