Question

Work out the coordinates of the points of intersection of the graphs

Answer 1

Step-by-step explanation:

The point of intersection occurs when the two graphs have equal values of  x and  y at the same time. There is only one solution, because two straight lines can only intersect once.

$y =4x^2 + 8x + 3 , y = x + 5\\\\4x^2 + 8x + 3 = x +5\\\\4x^2 + 8x + 3 -x -5 = 0\\\\4x^2 +7x -2 =0\\\\4x^2 + 8x -x - 2 = 0\\\\4x(x + 2) -1(x+2) = 0\\\\(4x - 1)(x+2) =0\\\\4x -1 = 0 , x+2 =0\\\\x = \frac{1}{4}, -2$

$x_1 = \frac{1}{4}, x_2 =-2$

$y_1 = 4x^2 + 8x + 3 \\\\x_1 = \frac{1}{4}\\\\y_1 = 4(\frac{1}{4})^2 + 8(\frac{1}{4}) + 3 = \frac{1}{4} + 2 + 3 = \frac{1}{4} +5 = \frac{21}{4}\\\\y_2 = x+ 5\\x_2 = -2\\y_2 = -2 + 5 =3$

Answer :

$(x_1, y_1) = (\frac{1}{4} , \frac{21}{4})\\\\(x_2,y_2) = (-2, 3)$

Answer 2

Answer:

(- 2, 3 ) and ($\frac{1}{4}$, $\frac{21}{4}$ )

Step-by-step explanation:

Given the 2 equations

y = 4x² + 8x + 3 → (1)

y = x + 5 → (2)

Substitute y = 4x² + 8x + 3 into (2)

4x² + 8x + 3 = x + 5 ( subtract x + 5 from both sides )

4x² + 7x - 2 = 0 ← in standard form

(x + 2)(4x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 2 = 0 ⇒ x = - 2

4x - 1 = 0 ⇒ 4x = 1 ⇒ x = $\frac{1}{4}$

Substitute these values into (2) for corresponding values of y

x = - 2 : y = - 2 + 5 = 3 ⇒ (- 2, 3 )

x = $\frac{1}{4}$ : y = $\frac{1}{4}$ + 5 = $\frac{21}{4}$ ⇒ ($\frac{1}{4}$, $\frac{21}{4}$ )