Q62. - Algebra 2

Question

Find the coordinates of the maximum or minimum value of each quadratic equation to the nearest hundredth.

 

$f\left(x\right)=7x^2+4x+1$

Step-by-Step Solution

Answer

The coordinates of the minimum value of the quadratic equation $f\left(x\right)=7x^2+4x+1$ to the nearest hundredth is $\left(-0.29,0.43\right)$.

Step 1. Given Information.

Given to determine the coordinates of the maximum or minimum value of the quadratic equation $f\left(x\right)=7x^2+4x+1$ to the nearest hundredth.

Step 2. Explanation .

The maximum or minimum value of a quadratic function lies at the vertex of the graph.

For an equation of the form $f\left(x\right)=a{x}^2+bx+c$:

if a>0, it is an upwards opening parabola and so has a minimum value

if a<0, it is a downwards opening parabola and so has a maximum value.

So, the given quadratic equation has a minimum value.

 

For an equation of the form $f\left(x\right)=a{x}^2+bx+c$, the x-coordinate of the vertex is given by $x=-\frac{b}{2a}$

Here for the given equation, a=7,b=4

Plugging the values in the equation:

 $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{\left(4\right)}{2\left(7\right)}\\ x=-\frac{4}{14}\\ x=-\frac{2}{7}\\ x\approx -0.29\end{array}$

Hence the x-coordinate of the vertex is x=-0.29.

 

So, the y-coordinate of the vertex can be obtained by plugging the value in the equation.

Plugging $x=-\frac{2}{7}$ in the equation:

 $\begin{array}{l}f\left(x\right)=7x^2+4x+1\\ f\left(-\frac{2}{7}\right)=7{\left(-\frac{2}{7}\right)}^2+4\left(-\frac{2}{7}\right)+1\\ f\left(-\frac{2}{7}\right)=7\left(\frac{4}{49}\right)+4\left(-\frac{2}{7}\right)+1\\ f\left(-\frac{2}{7}\right)=\frac{4}{7}-\frac{8}{7}+\frac{7}{7}\\ f\left(-\frac{2}{7}\right)=\frac{3}{7}\\ f\left(-\frac{2}{7}\right)\approx 0.43\end{array}$

Hence the coordinates of the vertex is $\left(-0.29,0.43\right)$.

Step 3. Conclusion .

The coordinates of the minimum value of the quadratic equation $f\left(x\right)=7x^2+4x+1$ to the nearest hundredth is $\left(-0.29,0.43\right)$.