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Question: How do you write a quadratic function in vertex form that has vertex \(\left( 1,2 \right)\) and pass...

How do you write a quadratic function in vertex form that has vertex (1,2)\left( 1,2 \right) and passes through point (2,4)\left( 2,4 \right) ?

Explanation

Solution

The general equation for a quadratic function in vertex form is given by, y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k .In this expression substitute the values of the vertex which is given in the question. Now we get an equation in y,a,x  y,a,x\; further we need to substitute the values of the given point which passes through it to find the value of a. Now write it all together to get the quadratic expression which is formally known as a parabola.

Complete step by step solution:
The general equation for a quadratic function in vertex form is given by, y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k
The given vertex coordinates are (1,2)\left( 1,2 \right)
Now substitute these coordinates in place of h,k  h,k\;
On substituting h=1;k=2h=1;k=2 in the general equation for a quadratic function in vertex form we get,
⇒y=a(x−1)2+2\Rightarrow y=a{{\left( x-1 \right)}^{2}}+2
Now we are left with an equation in three variables, which are y,a,x  y,a,x\;
We are also provided with another coordinate, a point that passes through this function which is (2,4)\left( 2,4 \right)
Substitute this coordinate in place of x,y  x,y\; to find the values of aa
Upon placing the values in the recent equation, we get,
⇒4=a(2−1)2+2\Rightarrow 4=a{{\left( 2-1 \right)}^{2}}+2
Now evaluate the expression.
⇒4=a(1)2+2\Rightarrow 4=a{{\left( 1 \right)}^{2}}+2
⇒4=a+2\Rightarrow 4=a+2
⇒a=2\Rightarrow a=2
Now since we have found out the value of aa , rewrite the equation for the function with the value of aa and the vertex coordinates.
⇒y=2(x−1)2+2\Rightarrow y=2{{\left( x-1 \right)}^{2}}+2

We can solve this further and evaluate but it is better to leave the expression in this form since it represents the coordinates of the vertex, the length of the latus rectum, and all other details of the parabola just by looking at it.
Hence the quadratic function in vertex form is given by, y=2(x−1)2+2y=2{{\left( x-1 \right)}^{2}}+2

Note: The given quadratic function is denoted by a parabola. Whenever there are any squares or roots in the expression, never solve them in the primary steps itself. Solve them in the end because there are most likely chances that they get cancelled out or simplified in the further processes.