Question

use your calculator to find the solutions and the maximum or minimum point of the quadratic equation.
$y = 2x^2 + 8x + 7$
\bigcirc max (-2,1); x has no solutions
\bigcirc min (-2,-1); x is approximately -2.7 x is approximately -1.3
\bigcirc max (6,1); x = 4.75, x = 6.25
\bigcirc min (1,-6); x = -1.75, x = 3.75

Explanation:

Step1: Analyze the quadratic function

For a quadratic function \( y = ax^2 + bx + c \), if \( a>0 \), the parabola opens upward, so it has a minimum point. Here, \( a = 2>0 \), so the function has a minimum. The x - coordinate of the vertex (minimum point) is given by \( x=-\frac{b}{2a} \). For \( y = 2x^2+8x + 7 \), \( a = 2 \), \( b = 8 \), so \( x=-\frac{8}{2\times2}=- 2 \).

Step2: Find the y - coordinate of the vertex

Substitute \( x=-2 \) into the function: \( y=2\times(-2)^2 + 8\times(-2)+7=2\times4-16 + 7=8 - 16 + 7=-1 \). So the minimum point is \( (-2,-1) \).

Step3: Solve the quadratic equation \( 2x^2+8x + 7 = 0 \)

Using the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 8 \), \( c = 7 \). First, calculate the discriminant \( \Delta=b^2-4ac=8^2-4\times2\times7=64 - 56 = 8 \). Then \( x=\frac{-8\pm\sqrt{8}}{2\times2}=\frac{-8\pm2\sqrt{2}}{4}=\frac{-4\pm\sqrt{2}}{2}\approx\frac{-4\pm1.414}{2} \).
For \( + \): \( x=\frac{-4 + 1.414}{2}=\frac{-2.586}{2}\approx - 1.293\approx - 1.3 \)
For \( - \): \( x=\frac{-4-1.414}{2}=\frac{-5.414}{2}\approx - 2.707\approx - 2.7 \)

Answer:

Min (-2,-1); x is approximately -2.7 x is approximately -1.3