Answer:
[tex]\textsf{27.} \quad \textsf{Vertex}:(-4,-5) \quad \textsf{$x$-intercepts}: x=-4+\sqrt{5}, \:\:x=-4-\sqrt{5}[/tex]
[tex]\textsf{28.} \quad \textsf{Vertex}:(-5,-11) \quad \textsf{$x$-intercepts}: x = -5+\sqrt{11}, \:\:x=-5-\sqrt{11}[/tex]
[tex]\textsf{29.} \quad \textsf{Vertex}:(1,16) \quad \textsf{$x$-intercepts}: x = 5, \:\:x=-3[/tex]
Step-by-step explanation:
Vertex form of a quadratic function:
[tex]\boxed{y=a(x-h)^2+k}[/tex]
where:
- (h, k) is the vertex.
- [tex]a[/tex] is some constant.
Question 27
Given function:
[tex]g(x)=x^2+8x+11[/tex]
Change to vertex form by completing the square.
Add and subtract the square of half the coefficient of x:
[tex]\implies g(x)=x^2+8x+11 +\left(\dfrac{8}{2}\right)^2-\left(\dfrac{8}{2}\right)^2[/tex]
[tex]\implies g(x)=x^2+8x+11 +16-16[/tex]
[tex]\implies g(x)=x^2+8x+16+11-16[/tex]
[tex]\implies g(x)=x^2+8x+16-5[/tex]
Factor the perfect trinomial:
[tex]\implies g(x)=(x+4)^2-5[/tex]
Therefore, the vertex is (-4, -5).
To find the x-intercepts, set the function to zero and solve for x:
[tex]\begin{aligned}g(x) & = 0\\\implies (x+4)^2 -5 & = 0\\(x+4)^2 & = 5\\\sqrt{(x+4)^2} & = \sqrt{5}\\x+4&=\pm\sqrt{5}\\x+4-4&=-4\pm\sqrt{5}\\x&=-4\pm\sqrt{5}\end{aligned}[/tex]
Therefore, the x-intercepts are:
[tex]x = -4+\sqrt{5}, \quad x = -4-\sqrt{5}[/tex]
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Question 28
Given function:
[tex]f(x)=x^2+10x+14[/tex]
Change to vertex form by completing the square.
Add and subtract the square of half the coefficient of x:
[tex]\implies f(x)=x^2+10x+14+\left(\dfrac{10}{2}\right)^2-\left(\dfrac{10}{2}\right)^2[/tex]
[tex]\implies f(x)=x^2+10x+14+25-25[/tex]
[tex]\implies f(x)=x^2+10x+25+14-25[/tex]
[tex]\implies f(x)=x^2+10x+25-11[/tex]
Factor the perfect trinomial:
[tex]\implies f(x)=(x+5)^2-11[/tex]
Therefore, the vertex is (-5, -11).
To find the x-intercepts, set the function to zero and solve for x:
[tex]\begin{aligned}f(x) & = 0\\\implies (x+5)^2 -11 & = 0\\(x+5)^2 & = 11\\\sqrt{(x+5)^2} & = \sqrt{11}\\x+5&=\pm\sqrt{11}\\x+5-5&=-5\pm\sqrt{11}\\x&=-5\pm\sqrt{11}\end{aligned}[/tex]
Therefore, the x-intercepts are:
[tex]x = -5+\sqrt{11}, \quad x = -5-\sqrt{11}[/tex]
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Question 29
Given function:
[tex]f(x)=-(x^2-2x-15)[/tex]
Change to vertex form by completing the square.
Add and subtract the square of half the coefficient of x:
[tex]f(x)=-\left(x^2-2x-15+\left(\dfrac{-2}{2}\right)^2-\left(\dfrac{-2}{2}\right)^2\right)[/tex]
[tex]f(x)=-\left(x^2-2x-15+1-1\right)[/tex]
[tex]f(x)=-\left(x^2-2x+1-15-1\right)[/tex]
[tex]f(x)=-\left(x^2-2x+1-16\right)[/tex]
Factor the perfect trinomial:
[tex]f(x)=-\left((x-1)^2-16\right)[/tex]
Simplify:
[tex]f(x)=-(x-1)^2+16[/tex]
Therefore, the vertex is (1, 16).
To find the x-intercepts, set the function to zero and solve for x:
[tex]\begin{aligned}f(x) & = 0\\\implies -(x-1)^2 +16 & = 0\\(x-1)^2 -16 & = 0\\(x-1)^2 & = 16\\\sqrt{(x-1)^2} & = \sqrt{16}\\x-1&=\pm4\\x-1+1&=1\pm4\\x&=5,-3\end{aligned}[/tex]
Therefore, the x-intercepts are:
[tex]x = 5, \quad x=-3[/tex]
