Question

Find the slope of the line passing through the vertex and the y-intercept of the quadratic function

Answer 1

Find y intercept

• y=5(0)²+20(0)-7
• y=0-7
• y=-7

Point(0,-7)

Find vertex

x coordinate

• -b/2a
• -20/10
• -2

y coordinate

• y=5(4)-40-7
• y=-27

Vertex at (-2,-27)

Slope

• m=(-27+7)/-2-0
• m=-20/-2
• m=10

Answer 2

Answer:

10

Step-by-step explanation:

Vertex

The x-coordinate of the vertex of a quadratic equation in the form

$f(x)=ax^2+bx+c\quad \textsf{is} \quad -\dfrac{b}{2a}$

Given function:

$f(x)=5x^2+20x-7$

$\implies a=5, \quad b=20, \quad c=-7$

x-coordinate of the vertex

$\implies -\dfrac{b}{2a}=-\dfrac{20}{2(5)}=-2$

To find the y-coordinate of the vertex, substitute the found value of x into the function:

$\begin{aligned}\implies f(-2) & =5(-2)^2+20(-2)-7\\& = 5(4)-40-7\\& = 20-47\\& = -27\end{aligned}$

Therefore, the coordinates of the vertex are (-2, -27).

y-intercept

The y-intercept is when the curve crosses the y-axis, so when x = 0.

To find the y-coordinate of the y-intercept, substitute x = 0 into the function:

$\begin{aligned}\implies f(0) & =5(0)^2+20(0)-7\\& = 0 + 0-7\\& = -7\end{aligned}$

Therefore, the coordinates of the y-intercept are (0, -7).

Slope

To find the slope of the line passing through the vertex and the y-intercept, simply substitute the found points into the slope formula:

$\implies \sf slope=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{-27-(-7)}{-2-0}=\dfrac{-20}{-2}=10$

Therefore, the slope of the line passing through the vertex and the y-intercept of the given quadratic function is 10.

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