What is the y-intercept of the quadratic function: f(x)=(4x-3) (2x-1)

Answers

Y intercept is (0, -22)
To find the smallest y-value, you substitute x=0 into each other the functions.

For function a(x), do a(0):
a(0) = (0+5)2-7 = 3

For function b(x), it gives you (0,6).

For function c(x), just look at where the line crosses the x-axis. It crosses it as -2, and there's a point at (0,-2).

So since -2 is the smallest, function c(x) has the smallest y-intercept: -2
I'm not sure which form they want the final eqn in but i hope this helps
Save me!  the factored form for a quadratic function is f(x)=(x‒p)(x‒q). the standard form for a qua
I think the answer will be A

a

Step-by-step explanation:

A.

Step-by-step explanation:

A quadratic function may only intercept the y-axis once. This is because the definition of a y-intercept tells us that it is the point where x = 0. If there are multiple points where x = 0, then the expression cannot be a function. As such, a quadratic function can only have one.

Choice B contradicts the previous statement.

Choice C is wrong because the x-intercept is the zero of the function. This is because at a given x-coordinate that touches the x-axis, the y-coordinate will be 'zero'.

Choice D is wrong because the y-intercept is where x = 0.

I hope this helps!

a. A quadratic can only have one y-intercept.

Step-by-step explanation:

1) x=\dfrac{5}{2}

2)  t_1=-1 and t_2=6

3) (0,12)

4) \left(\dfrac{5}{2},\dfrac{49}{2}\right)

Step-by-step explanation:

The function h(t) =-2t^2 + 10t + 12 represents the height (h) of the ball after t seconds.

Find the vertex of the parabola:

t_v=-\dfrac{b}{2a}=-\dfrac{10}{2\cdot (-2)}=\dfrac{5}{2}\\ \\h_v=h\left(\dfrac{5}{2}\right)=-2\cdot\left(\dfrac{5}{2}\right)^2+10\cdot \dfrac{5}{2}+12=-2\cdot \dfrac{25}{4}+25+12=-\dfrac{25}{2}+37=\dfrac{49}{2}

Hence, the vertex of the function is at point \left(\dfrac{5}{2},\dfrac{49}{2}\right).

The axis of symmetry of the function is vertical line which passes through the vertex, so its equation is

x=\dfrac{5}{2}

To find y-intercept, equat t to 0 and find h:

h=-2\cdot 0^2+10\cdot 0+12=12

Hence, y-intercept is at point (0,12)

To find the roots of the quadratic finction, equate h to 0 and solve the equation for t:

h=0\Rightarrow -2t^2+10t+12=0\\ \\t^2-5t-6=0\ [\text{Divided by -2}]\\ \\D=(-5)^2-4\cdot 1\cdot (-6)=25+24=49\\ \\t_{1,2}=\dfrac{-(-5)\pm \sqrt{49}}{2\cdot 1}=\dfrac{5\pm 7}{2}=6,\ -1

Therefore, two roots are t_1=-1 and t_2=6

x= 1/2 or x= 3/4

Step-by-step explanation:

x= 1/2 or x= 3/4

Step-by-step explanation:



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