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Question: How do you write \(F\left( x \right) = {x^2} - 2x + 5\) in vertex form?...

How do you write F(x)=x22x+5F\left( x \right) = {x^2} - 2x + 5 in vertex form?

Explanation

Solution

We have to rewrite the given equation in vertex form. For this, complete the square for x22x+5{x^2} - 2x + 5 and use the form ax2+bx+ca{x^2} + bx + c, to find the values of aa, bb, and cc. Consider the vertex form of a parabola, a(x+d)2+ea{\left( {x + d} \right)^2} + e. Next, substitute the values of aa and bb into the formula d=b2ad = \dfrac{b}{{2a}} and simplify the right side. Next, find the value of ee using the formula e=cb24ae = c - \dfrac{{{b^2}}}{{4a}}. Next, substitute the values of aa, dd, and ee into the vertex form a(x+d)2+ea{\left( {x + d} \right)^2} + e. Next, set yy equal to the new right side and get the required vector form.

Formula used:
Vertex form of a parabola: a(x+d)2+ea{\left( {x + d} \right)^2} + e
d=b2ad = \dfrac{b}{{2a}}
e=cb24ae = c - \dfrac{{{b^2}}}{{4a}}
Vertex form: y=a(xh)2+ky = a{\left( {x - h} \right)^2} + k
Vertex: (h,k)\left( {h,k} \right)
p=14ap = \dfrac{1}{{4a}}
Focus: (h,k+p)\left( {h,k + p} \right)
Directrix: y=kpy = k - p

Complete step by step solution:
We have to rewrite the given function in vertex form.
For this, complete the square for x22x+5{x^2} - 2x + 5.
Use the form ax2+bx+ca{x^2} + bx + c, to find the values of aa, bb, and cc.
a=1,b=2,c=5a = 1,b = - 2,c = 5
Consider the vertex form of a parabola.
a(x+d)2+ea{\left( {x + d} \right)^2} + e
Now, substitute the values of aa and bb into the formula d=b2ad = \dfrac{b}{{2a}}.
d=22×1d = \dfrac{{ - 2}}{{2 \times 1}}
Simplify the right side.
d=1\Rightarrow d = - 1
Find the value of ee using the formula e=cb24ae = c - \dfrac{{{b^2}}}{{4a}}.
e=5(2)24×1e = 5 - \dfrac{{{{\left( { - 2} \right)}^2}}}{{4 \times 1}}
e=4\Rightarrow e = 4
Now, substitute the values of aa, dd, and ee into the vertex form a(x+d)2+ea{\left( {x + d} \right)^2} + e.
(x1)2+4{\left( {x - 1} \right)^2} + 4
Set yy equal to the new right side.
y=(x1)2+4y = {\left( {x - 1} \right)^2} + 4
Which is the required vertex form.

y=(x1)2+4y = {\left( {x - 1} \right)^2} + 4 is the vertex form of a given function.

Note: Vertex form of the quadratic function:
y=a(xh)2+ky = a{\left( {x - h} \right)^2} + k where (h,k)\left( {h,k} \right) is the vertex or the “center” of the quadratic function or the parabola.
We can also find the vertex form of a given function by plotting the function and determining the vertex, (h,k)\left( {h,k} \right) of the given function.

From the graph, we can observe that (1,4)\left( {1,4} \right) is the vertex of the parabola.
Put the value of a,h,ka,h,k in y=a(xh)2+ky = a{\left( {x - h} \right)^2} + k.
y=(x1)2+4y = {\left( {x - 1} \right)^2} + 4 (As a=1a = 1)