Question

How do you find the vertex of \[y = 4{x^2} + 8x + 7\]?

Answer 1

Hint: We compare the given quadratic equation with general quadratic equation and write values of coefficients. Use the formula to calculate the value of vertex and substitute the values in it.
* The general quadratic equation is \[a{x^2} + bx + c = 0\]
* Vertex \[(h,k)\] is given by the formula \[h = - \dfrac{b}{{2a}};k = f(h)\], where f is the function given to us.

Complete step-by-step answer:
We are given the equation \[y = 4{x^2} + 8x + 7\]
We compare the equation on right hand side to the general quadratic equation \[a{x^2} + bx + c = 0\]
We get the values \[a = 4,b = 8,c = 7\]
We know that vertex \[(h,k)\]is given by the formula \[h = - \dfrac{b}{{2a}};k = f(h)\], where f is the function given to us
Here function is \[y = 4{x^2} + 8x + 7\] … (1)
We substitute the values of ‘a’ and ‘b’ in formula of ‘h’
\[ \Rightarrow h = - \dfrac{8}{{2 \times 4}}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow h = - 1\]
Now we calculate the value of ‘k’ by substituting value of ‘h’ in place of ‘x’ in the function
\[ \Rightarrow k = f( - 1)\]
\[ \Rightarrow k = 4{( - 1)^2} + 8( - 1) + 7\]
Calculate the products on right hand side of the equation
\[ \Rightarrow k = 4 - 8 + 7\]
Add the positive terms on right hand side of the equation
\[ \Rightarrow k = 11 - 8\]
Calculate the difference on right hand side of the equation
\[ \Rightarrow k = 3\]
So, we get the value of h as -1 and k as 3
\[ \Rightarrow \]Vertex \[(h,k) = ( - 1,3)\]

\[\therefore \] The vertex of \[y = 4{x^2} + 8x + 7\] is \[( - 1,3)\]

Note:
Many students make the mistake of writing the value of k wrong as they get confused of what the function is. Keep in mind here y is a function of x i.e. we have to substitute the value of x on the right side and calculate the value of y. Students get confused and many times write the value of y as h and calculate the roots of the obtained equation which is wrong.