Question

Will give brainliest and 40 points. Just last question C needed, c) Use the graph to find the two x values.

Answer 1

Answer:

a) Complete the table of values for $y=x^2+x-4$

$y=(-3)^2+(-3)-4\\y=9-7\\y=2$

$y=(-2)^2+(-2)-4\\y=4-6\\y=-2$

$y=(-1)^2+(-1)-4\\y=1-5\\y=-4$

$y=(0)^2+(0)-4\\y=-4$

$y=(1)^2+(1)-4\\y=1-3\\y=-2$

$y=(2)^2+(2)-4\\y=4-2\\y=2$

$y=(3)^2+(3)-4\\y=9-1\\y=8$

b)

$y=x^2+x-4$

x-intercepts is when y = 0.

$0=x^2+x-4$

Using Quadratic Formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-4\right)}}{2\cdot \:1}$

$x=\frac{-1+\sqrt{17}}{2\cdot \:1}$

$x_{1}=\frac{-1+\sqrt{17}}{2}$

$\:x_{2}=\frac{-1-\sqrt{17}}{2}$

Taking $x_{1}=\frac{-1+\sqrt{17}}{2}$

$x_{1}\approx 1.56$

Therefore it is B.

c)

Approximations:

$x_{1}\approx 1.56$

$x_{2}\approx -2.56$

Exact values:

$x_{1}=\frac{-1+\sqrt{17}}{2}$

$\:x_{2}=\frac{-1-\sqrt{17}}{2}$