Answer:
see explanation
Step-by-step explanation:
Given a quadratic in standard form , y = ax² + bx + c ( a ≠ 0 ), then
The x- coordinate of the vertex, which is also the equation of the axis of symmetry is
[tex]x_{vertex}[/tex] = - [tex]\frac{b}{2a}[/tex]
y = x² + 12x + 32 ← is in standard form
with a = 1, b = 12 , then
[tex]x_{vertex}[/tex] = - [tex]\frac{12}{2}[/tex] = - 6
Substitute x = - 6 into y for corresponding y- coordinate
y = (- 6)² + 12(- 6) + 32 = 36 - 72 + 32 = - 4
Thus
equation of axis of symmetry is x = - 6
vertex = (- 6, - 4 )
To find the y- intercept , let x = 0
y = 0² + 12(0) + 32 = 32 ← y- intercept
The axis of symmetry is x = -6, vertex is (-6,-4) and y-intercept is 32.
Important information:
Substitute [tex]x=0[/tex] to find the y-intercept.
[tex]y=(0)^2+12(0)+32[/tex]
[tex]y=32[/tex]
The vertex form of a quadratic function is:
[tex]y=a(x-h)^2+k[/tex] ...(i)
Where a is constant, (h,k) is vertex and x=h is the axis of symmetry.
Adding and subtracting the square of half of coefficient of x, the given quadratic function can be written as:
[tex]y=x^2+12x+32+6^2-6^2[/tex]
[tex]y=(x^2+12x+6^2)+32-36[/tex]
[tex]y=(x+6)^2-4[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]a=1,h=-6,k=-4[/tex]
Therefore, the axis of symmetry is x = -6, vertex is (-6,-4) and y-intercept is 32.
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