Question

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

Answer 1

$(-3.07,0),(0.57,0)$

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

$\begin{gathered} for \\ ax^2+bx+c=0 \\ the\text{ solution for x is } \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}$

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

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

therefore,

$\begin{gathered} x_1=\frac{-10+14.56}{8}=0.57 \\ x_1=\frac{-10-14.56}{8}=-3.07 \end{gathered}$

so, the coordinates are:

$(-3.07,0),(0.57,0)$

I hope this helps you

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