Question

Please help me with the last part of the question, it keeps showing up as wrong?

Answer 1

`(-3.07,0),(0.57,0)`

Step-by-step explanation

Step 1

given the function:

`f(x)=4x^2+10x-7`

we know that the x-intercept is the x-coordinate of a point where the parabola intersects the x-axis, this happens when y =0

so, we need to set the function equals zero and solve for x

a) use the quadratic formula

`\begin{gathered} for \\ ax^2+bx+c=0 \\ the\text{ solution for x is } \\ x=(-b\pm√(b^2-4ac))/(2a) \end{gathered}`

so

let

`\begin{gathered} 4x^2+10x-7=0\text{ \lparen}set\text{ the function equals 0\rparen} \\ now,\text{ } \\ 4x^2+10x-7=ax^2+bx+c \\ so \\ a=4 \\ b=10 \\ c=-7 \end{gathered}`

now, replace in the quadratic formula

`\begin{gathered} x=(-b\pm√(b^2-4ac))/(2a) \\ x=(-(10)\pm√(10^2-4(4)(-7)))/(2(4)) \\ x=(-10\pm√(212))/(8) \\ x=(-10\pm14.56)/(8) \end{gathered}`

therefore,

`\begin{gathered} x_1=(-10+14.56)/(8)=0.57 \\ x_1=(-10-14.56)/(8)=-3.07 \end{gathered}`

so, the coordinates are:

`(-3.07,0),(0.57,0)`

I hope this helps you

.