Answer:
Factored form: y = (x+1)(x-8)
x-intercept: (-1, 0) and (8, 0)
Axis of symmetry: x = 7/2
Vertex: (7/2, -81/4)
Domain: All real numbers
Range: y ≥ -81/4
Explanations:
Given the quadratic equation expressed as:
[tex]y=x^2-7x-8[/tex]
Factorize
[tex]\begin{gathered} y=x^2-8x+x-8 \\ y=x(x-8)+1(x-8) \\ y=(x+1)(x-8)\text{ Factored form} \end{gathered}[/tex]
The x-intercept is the point where y= 0. Substitute y = 0 into the factored form
[tex]\begin{gathered} (x+1)(x-8)=0 \\ x=-1\text{ }and\text{ }8 \\ The\text{ x-intercept are \lparen-1, 0\rparen and \lparen8, 0\rparen} \end{gathered}[/tex]
The axis of symmetry of the equation is given as x = -b/2a where:
a = 1
b = -7
Substitute:
[tex]\begin{gathered} axis\text{ of symmetry:}x=\frac{-(-7)}{2(1)} \\ axis\text{ of symmetry: }x=\frac{7}{2} \end{gathered}[/tex]
The vertex form of the equation is in the form (x-h)^2+k where (h, k) is the vertex. Rewrite in vertex form:
[tex]\begin{gathered} y=x^2-7x-8 \\ y=x^2-7x+(-\frac{7}{2})^2-(-\frac{7}{2})^2-8 \\ y=(x-\frac{7}{2})^2-\frac{49}{4}-8 \\ y=(x-\frac{7}{2})^2-\frac{81}{4} \end{gathered}[/tex]
The vertex of the function will be (7/2, -81/4)
The domain are the independent values of the function for which it exists. The domain of the given quadratic function exists on all real number that is:
[tex]Domain:(-\infty,\infty)[/tex]
The range of the function are the dependent value for which it exist. For the given function, the range is given as:
[tex]Range:[-\frac{81}{4},\infty)[/tex]
