is noteworthy that the leading term has a negative coefficient, meaning this parabola is opening downwards like a "camel hump", so it reaches a maximum point and then goes back down, and of course the maximum point is at its vertex.
[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ H(x)=\stackrel{\stackrel{a}{\downarrow }}{-\frac{1}{2}}x^2\stackrel{\stackrel{b}{\downarrow }}{+4}x\stackrel{\stackrel{c}{\downarrow }}{-5} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{4}{2\left( -\frac{1}{2} \right)}~~,~~-5-\cfrac{4^2}{4\left( -\frac{1}{2} \right)} \right)\implies \left( 4~,~-5+\cfrac{16}{2} \right)\implies (4~~,~~3)[/tex]
Answer:
(4, -3)
Step-by-step explanation:
I'm assuming that you meant H(x) = -(1/2)x^2 + 4x - 5. Here, the coefficients of this quadratic are a = -1/2, b = 4 and c = -5.
The axis of symmetry is x = -b/(2a). This axis goes through the vertex. Here the axis of symmetry is x = -(4) / [ 2(-1/2) ], or x = 4.
Evaluating H(x) at x = 4 gives us the y value of the vertex. It is:
H(4) = (-1/2)(4)^2 + 4(4) - 5, or H(4) = -8 + 16 - 5, or 3.
We know that this function must have a max because a is - and therefore the graph opens down.
The vertex and the maximum is (4, 3).
