Answer:
Minimum = (-3, -2)
Step-by-step explanation:
Standard form of a quadratic function:
[tex]f(x)=ax^2+bx+c[/tex]
If a > 0 the parabola opens upwards and the curve has a minimum point.
If a < 0 the parabola opens downwards and curve has a maximum point.
Given function:
[tex]f(x)=x^2+6x+7[/tex]
As a > 0, the parabola opens upwards and so the curve has a minimum point.
The minimum/maximum point of a quadratic function is its vertex.
Vertex form of a quadratic function:
[tex]f(x)=(x-h)^2+k[/tex]
Where (h, k) is the vertex.
To rewrite the given function in vertex form, complete the square.
Add and subtract the square of half the coefficient of the term in x:
[tex]\implies f(x)=x^2+6x+7 +\left(\dfrac{6}{2}\right)^2-\left(\dfrac{6}{2}\right)^2[/tex]
[tex]\implies f(x)=x^2+6x+7 +9-9[/tex]
[tex]\implies f(x)=x^2+6x+9+(7 -9)[/tex]
[tex]\implies f(x)=x^2+6x+9-2[/tex]
Factor the perfect square trinomial formed by x²+6x+9:
[tex]\implies f(x)=(x+3)^2-2[/tex]
Compare with the vertex form:
Therefore, the vertex is (-3, -2) and so the minimum value of the given function is (-3, -2).